{"id":"403","title":"Eddy Current Level 2 Practice Questions 2026: Free Test","slug":"et-level-2-practice-questions-2026-free-mock-exam","date":"June 24, 2026","author":"Anoop Rayavarapu","category":"Practice Questions","metaDescription":"30 original eddy current (ET) Level II practice questions with answers: skin depth, probes, fill factor, impedance plane, tubing, standards, limitations.","snippet":"30 original eddy current (ET) Level II practice questions with answers: skin depth, probes, fill factor, impedance plane, tubing, standards, limitations.","content":"\n<p class=\"ptq-meta\" style=\"color:#475569;margin:0 0 16px;\">30 original questions in 6 topics, each with the correct answer, an explanation and its basis. Last reviewed October 16, 2026.</p>\n<section data-practice-test=\"2026-10\" aria-label=\"About this practice test\"><h2>About this ET Level II practice test</h2><p>These 30 multiple-choice questions are original study questions written by Atlantis NDT for eddy current technicians preparing for a Level II examination. They are not ASNT exam questions. They follow the Level II job: select probes and frequency, standardize on reference standards, read the impedance plane, carry out surface and tubing examinations, evaluate indications against a procedure and report the results.</p><p>Under ASNT SNT-TC-1A (2024 edition), the employer certifies Level II technicians after general, specific and practical examinations set out in its written practice. ASNT&apos;s central ASNT NDT Level II examinations do not currently include eddy current testing, so ET certification rests on the employer&apos;s own examinations. Use this set to find weak areas before them.</p><p>Atlantis NDT runs NDT training year-round, with theory online and hands-on practice on its Practical NDT simulator or in person at the client&apos;s own facility. Simulator practice supports training; it does not replace the practical examination or the experience hours the employer&apos;s written practice requires.</p></section>\n<section aria-label=\"How to use this practice test\"><h2>How to use this practice test</h2><ul><li>Answer each question before opening the explanation.</li><li>For the depth-of-penetration questions, check which unit convention the formula uses before you substitute numbers.</li><li>Sketch the impedance plane when a question asks about signal direction or phase; it is easier to reason from a picture.</li><li>Note the topics where you miss questions, study them against your employer&apos;s procedure, then re-test.</li></ul><p style=\"margin:8px 0 0;\">Topics in this set:</p><ul><li><a href=\"#topic-principles-and-electromagnetic-theory\">Principles and electromagnetic theory</a> (6 questions)</li><li><a href=\"#topic-probes-coils-and-instrumentation\">Probes, coils and instrumentation</a> (5 questions)</li><li><a href=\"#topic-standardization-and-reference-standards\">Standardization and reference standards</a> (5 questions)</li><li><a href=\"#topic-impedance-plane-interpretation\">Impedance-plane interpretation</a> (6 questions)</li><li><a href=\"#topic-applications-and-techniques\">Applications and techniques</a> (4 questions)</li><li><a href=\"#topic-codes-procedures-and-limitations\">Codes, procedures and limitations</a> (4 questions)</li></ul></section>\n<section class=\"ptq-note\" aria-label=\"Practice questions, not exam questions\" style=\"border-left:4px solid #d97706;padding:8px 0 8px 16px;margin:24px 0;\"><h2>Practice questions, not exam questions</h2><p>Atlantis NDT wrote these questions as original study material. They are not taken from ASNT or from any employer&apos;s examination, and real examinations differ in wording, emphasis and format.</p><p>Certification is a separate step: under ASNT SNT-TC-1A the employer certifies its technicians after the examinations its written practice sets, and ASNT runs central Level II examinations for several methods. A good score here is a study signal, not a qualification.</p></section>\n<aside data-course-next-step=\"2026-10\" data-placement=\"answer\" aria-label=\"Next step\" style=\"border-left:4px solid #2563eb;padding:10px 0 10px 16px;margin:24px 0;\"><p style=\"margin:0\"><strong>Next step:</strong> Preparing for the ET Level 2 exam? The <a href=\"/et-level-2-training\" style=\"color:#2563eb;text-decoration:underline;\">ET Level 2 course</a> adds structured classroom and hands-on training. Employers: <a href=\"/consulting/ndt-consulting-level-iii\" style=\"color:#2563eb;text-decoration:underline;\">ASNT Level III consulting</a> handles procedures, the written practice and qualification exams.</p></aside>\n<section class=\"ptq-topic\" id=\"topic-principles-and-electromagnetic-theory\" aria-label=\"Principles and electromagnetic theory\"><h2>Principles and electromagnetic theory</h2><div class=\"ptq-question\" id=\"et2-01\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 1.</span> How are eddy currents produced in a test part?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> By the static field of a permanent magnet held near the part</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> By induction from the alternating magnetic field of an AC coil</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> By passing direct current through the part between two contacts</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> By ultrasonic vibration of the part surface</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: B. By induction from the alternating magnetic field of an AC coil</p><p style=\"margin:6px 0;\">An AC coil produces a changing magnetic field that induces circulating currents in a nearby conductor. A static field or direct current does not induce continuous eddy currents in a stationary part.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (electromagnetic induction)</p></details></div><div class=\"ptq-question\" id=\"et2-02\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 2.</span> What effect do the induced eddy currents have on the test coil?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> They set up an opposing secondary field that changes the coil&apos;s impedance</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> They raise the frequency of the alternating current in the coil</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> They reinforce the coil&apos;s primary field and leave its impedance unchanged</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> They only heat the part slightly and cannot be detected by the coil</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: A. They set up an opposing secondary field that changes the coil&apos;s impedance</p><p style=\"margin:6px 0;\">By Lenz&apos;s law the eddy-current field opposes the change that created it. Changes in that field, caused by flaws, conductivity, permeability or lift-off, alter the coil&apos;s resistance and inductive reactance, which the instrument displays.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (Lenz&apos;s law)</p></details></div><div class=\"ptq-question\" id=\"et2-03\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 3.</span> The standard depth of penetration (&#x3B4;) is the depth at which the eddy current density has fallen to about what fraction of its surface value?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> 50% (1/2)</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> 5% (1/20)</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> 37% (1/e)</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> 10% (1/10)</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: C. 37% (1/e)</p><p style=\"margin:6px 0;\">At one standard depth the density is 1/e, about 37% of the surface value. At about three standard depths it falls to about 5%, which is often used as the practical (effective) depth of penetration.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (skin effect)</p></details></div><div class=\"ptq-question\" id=\"et2-04\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 4.</span> Using &#x3B4; (mm) = 50&#xB7;&#x221A;(&#x3C1;/(f&#xB7;&#xB5;r)), with &#x3C1; in &#xB5;&#x3A9;&#xB7;cm and f in Hz, what is the standard depth of penetration in a non-magnetic aluminum alloy with &#x3C1; = 4.0 &#xB5;&#x3A9;&#xB7;cm at 10 kHz?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> 0.1 mm</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> 0.5 mm</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> 2.0 mm</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> 1.0 mm</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: D. 1.0 mm</p><p style=\"margin:6px 0;\">&#x3B4; = 50 &#xD7; &#x221A;(4.0 &#xF7; (10,000 &#xD7; 1)) = 50 &#xD7; &#x221A;0.0004 = 50 &#xD7; 0.02 = 1.0 mm, with &#xB5;r = 1 for a non-magnetic material. The constant 50 corresponds to 503 in the SI form &#x3B4; (m) = 503&#xB7;&#x221A;(&#x3C1;/(f&#xB7;&#xB5;r)) with &#x3C1; in &#x3A9;&#xB7;m.</p><p style=\"margin:6px 0;color:#475569;\">Basis: Standard depth of penetration equation</p></details></div><div class=\"ptq-question\" id=\"et2-05\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 5.</span> In a non-magnetic alloy, the standard depth of penetration is 1.0 mm at 10 kHz. What is it at 40 kHz?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> 0.25 mm</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> 2.0 mm</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> 4.0 mm</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> 0.5 mm</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: D. 0.5 mm</p><p style=\"margin:6px 0;\">&#x3B4; is proportional to 1/&#x221A;f. Raising the frequency by a factor of 4 reduces &#x3B4; by &#x221A;4 = 2, giving 0.5 mm. A fourfold change in &#x3B4; would need a sixteenfold change in frequency.</p><p style=\"margin:6px 0;color:#475569;\">Basis: Standard depth of penetration equation</p></details></div><div class=\"ptq-question\" id=\"et2-06\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 6.</span> Relative to the surface, what is the phase lag of the eddy currents at one standard depth of penetration?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> About 57&#xB0; (1 radian)</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> 0&#xB0; (no lag)</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> 90&#xB0; (a quarter cycle)</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> 180&#xB0; (half a cycle)</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: A. About 57&#xB0; (1 radian)</p><p style=\"margin:6px 0;\">Phase lag increases linearly with depth as x/&#x3B4; radians, so at one standard depth it is 1 radian, about 57&#xB0;. This phase lag is what lets signals from flaws at different depths be told apart by their phase.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (phase lag)</p></details></div></section><section class=\"ptq-topic\" id=\"topic-probes-coils-and-instrumentation\" aria-label=\"Probes, coils and instrumentation\"><h2>Probes, coils and instrumentation</h2><div class=\"ptq-question\" id=\"et2-07\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 7.</span> Compared with an absolute probe, how does a differential probe (two coils comparing adjacent areas) behave?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> It is more sensitive to gradual changes such as slow wall thinning and conductivity drift</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> It cannot detect small cracks because the two coils cancel</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> It largely ignores gradual changes and drift but responds well to small, abrupt flaws</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> It can only be used for conductivity measurement</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: C. It largely ignores gradual changes and drift but responds well to small, abrupt flaws</p><p style=\"margin:6px 0;\">A differential probe responds to the difference between its two coils, so changes common to both, such as gradual wall loss, slow conductivity variation or drift, largely cancel. Absolute probes are better for gradual changes and conductivity measurement.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (probe types)</p></details></div><div class=\"ptq-question\" id=\"et2-08\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 8.</span> Why is a standard bobbin coil insensitive to circumferential cracks in tubing?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Bobbin coils cannot operate at the high frequencies cracks need</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Circumferential cracks are always too shallow to give a signal</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> Circumferential cracks only occur in ferromagnetic tubes, which need other methods</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Its currents circulate parallel to the crack, so the crack barely disturbs them</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: D. Its currents circulate parallel to the crack, so the crack barely disturbs them</p><p style=\"margin:6px 0;\">A flaw is detected when it interrupts the eddy-current path. Bobbin-coil currents circulate around the tube, so axial cracks interrupt them strongly while circumferential cracks lie along the current path. Rotating or array probes are used for circumferential cracking.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (bobbin coils)</p></details></div><div class=\"ptq-question\" id=\"et2-09\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 9.</span> A bobbin probe with a coil diameter of 16.0 mm is used in a tube with an inside diameter of 18.0 mm. What is the fill factor?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> 89%</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> 79%</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> 63%</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> 94%</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: B. 79%</p><p style=\"margin:6px 0;\">For an internal (bobbin) coil the fill factor is (coil diameter &#xF7; tube inside diameter)&#xB2; = (16.0 &#xF7; 18.0)&#xB2; = 0.889&#xB2; &#x2248; 0.79, or 79%. Using the diameter ratio without squaring gives the wrong value of 89%.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (fill factor)</p></details></div><div class=\"ptq-question\" id=\"et2-10\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 10.</span> A flat pancake (surface) probe is placed on a plate. Which discontinuity is it least likely to detect?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> A surface crack perpendicular to the surface</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> A pit open to the surface</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> A surface-breaking fatigue crack running across the probe path</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> A lamination lying parallel to the surface</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: D. A lamination lying parallel to the surface</p><p style=\"margin:6px 0;\">The eddy currents from a surface probe flow in loops parallel to the surface. A lamination parallel to the surface lies in the plane of the current flow and causes little disturbance, whereas cracks perpendicular to the surface interrupt the currents.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (current flow and flaw orientation)</p></details></div><div class=\"ptq-question\" id=\"et2-11\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 11.</span> What is the main advantage of a shielded (or ferrite-focused) pencil probe for inspection near edges, fasteners or geometry changes?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> It concentrates the field, reducing edge effect and nearby-feature signals</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> It increases the standard depth of penetration directly under the probe tip</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> It allows testing of non-conductive materials such as plastics</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> It removes the need for a reference standard near fasteners</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: A. It concentrates the field, reducing edge effect and nearby-feature signals</p><p style=\"margin:6px 0;\">Shielding or a ferrite core concentrates the field under the probe tip, so the probe can work closer to edges and fasteners before their signals swamp a flaw response. It does not increase penetration or make non-conductors testable.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET practice (probe selection)</p></details></div></section><section class=\"ptq-topic\" id=\"topic-standardization-and-reference-standards\" aria-label=\"Standardization and reference standards\"><h2>Standardization and reference standards</h2><div class=\"ptq-question\" id=\"et2-12\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 12.</span> Why should an eddy current reference standard be made, as far as practical, from the same material, heat-treat condition and dimensions as the part under test?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Because the standard is used only to check the probe cable and connectors</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Because the coil response depends on conductivity, permeability and geometry</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> Because codes forbid artificial flaws in standards made from other materials</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Because the instrument gain cannot be adjusted between different materials</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: B. Because the coil response depends on conductivity, permeability and geometry</p><p style=\"margin:6px 0;\">Signal amplitude and phase from a given flaw depend on the material&apos;s electrical and magnetic properties and on the geometry. A mismatched standard gives a misleading set-up for amplitude, phase and depth estimation.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET practice; ASME BPVC Section V, Article 8</p></details></div><div class=\"ptq-question\" id=\"et2-13\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 13.</span> Why is the impedance-plane display usually rotated so that the lift-off signal runs horizontally?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> So flaw signals, out of phase with lift-off, show mainly as vertical deflections</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Because the instrument cannot display the lift-off signal in the vertical direction</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> To increase the standard depth of penetration at the test frequency</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> To switch the instrument from absolute to differential mode for crack detection</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: A. So flaw signals, out of phase with lift-off, show mainly as vertical deflections</p><p style=\"margin:6px 0;\">Setting lift-off horizontal is a phase adjustment. With probe wobble along the horizontal axis, signals with a different phase, such as cracks, show mainly as vertical components, and the vertical channel or an alarm can then be used with less interference from lift-off.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET practice (phase set-up)</p></details></div><div class=\"ptq-question\" id=\"et2-14\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 14.</span> In bobbin-coil examination of non-ferromagnetic tubing, how is the depth of an outside-surface flaw commonly estimated?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> From the signal duration on the strip chart at the probe speed used</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> From the probe&apos;s travel speed when the signal appears</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> By reading its phase against a phase-depth curve from reference flaws</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> From the fill factor alone</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: C. By reading its phase against a phase-depth curve from reference flaws</p><p style=\"margin:6px 0;\">Reference standards with machined flaws of known depth, such as through-wall and flat-bottom holes, give a curve of phase angle against depth at the test frequency. The phase of a field signal is read against that curve; ASME Section V Article 8 uses this approach for tubing.</p><p style=\"margin:6px 0;color:#475569;\">Basis: ASME BPVC Section V, Article 8</p></details></div><div class=\"ptq-question\" id=\"et2-15\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 15.</span> Using %IACS = 172.41 &#xF7; &#x3C1; (with &#x3C1; in &#xB5;&#x3A9;&#xB7;cm), what is the conductivity of a material with a resistivity of 3.45 &#xB5;&#x3A9;&#xB7;cm?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> About 29% IACS</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> About 50% IACS</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> About 59% IACS</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> About 100% IACS</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: B. About 50% IACS</p><p style=\"margin:6px 0;\">172.41 &#xF7; 3.45 &#x2248; 50% IACS. For comparison, 100% IACS corresponds to annealed copper at about 1.724 &#xB5;&#x3A9;&#xB7;cm.</p><p style=\"margin:6px 0;color:#475569;\">Basis: IACS conductivity conversion</p></details></div><div class=\"ptq-question\" id=\"et2-16\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 16.</span> During conductivity measurement of aluminum parts, why must the temperature of the part and the conductivity standards be considered?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Because conductivity varies with temperature, causing measurement error</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Because the probe frequency changes with room temperature</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> Because aluminum becomes ferromagnetic when cold</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Because temperature has no effect and is recorded only for information</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: A. Because conductivity varies with temperature, causing measurement error</p><p style=\"margin:6px 0;\">Electrical conductivity of metals falls as temperature rises. Measurements are made with the part and standards at similar temperatures, or corrected, as the procedure requires.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET practice (conductivity measurement)</p></details></div></section><section class=\"ptq-topic\" id=\"topic-impedance-plane-interpretation\" aria-label=\"Impedance-plane interpretation\"><h2>Impedance-plane interpretation</h2><div class=\"ptq-question\" id=\"et2-17\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 17.</span> On a normalized impedance plane, how does the response when an absolute coil is placed on ferromagnetic carbon steel differ from placing it on aluminum?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Both move in the same direction by the same amount</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> On steel the coil impedance does not change at all</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> On aluminum the inductive reactance rises above the air point, and on steel it falls</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> On steel, permeability raises the reactance; on aluminum, the reactance falls</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: D. On steel, permeability raises the reactance; on aluminum, the reactance falls</p><p style=\"margin:6px 0;\">High permeability concentrates the coil&apos;s magnetic flux and raises inductive reactance, which outweighs the eddy-current effect in most steels. In a non-magnetic conductor the opposing eddy-current field lowers the inductive reactance.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (impedance plane)</p></details></div><div class=\"ptq-question\" id=\"et2-18\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 18.</span> On the normalized impedance plane of a non-ferromagnetic material, which change moves the operating point along the conductivity curve in the same direction as an increase in conductivity?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> A decrease in test frequency</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> An increase in lift-off</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> An increase in test frequency</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> A decrease in coil diameter</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: C. An increase in test frequency</p><p style=\"margin:6px 0;\">The position on the curve depends on the product of frequency, conductivity and permeability, together with coil size. Raising the frequency therefore moves the point the same way as raising the conductivity. Lift-off moves the point along a different path, toward the air point.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (impedance plane)</p></details></div><div class=\"ptq-question\" id=\"et2-19\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 19.</span> Compared with a surface-breaking crack, how does the signal from a similar flaw located below the surface of a non-magnetic material typically differ?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> It is larger and has the same phase as the surface signal</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> It has smaller amplitude and a greater phase lag</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> It is identical, since depth does not affect eddy current signals</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> It always appears exactly along the lift-off line</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: B. It has smaller amplitude and a greater phase lag</p><p style=\"margin:6px 0;\">Eddy-current density decreases and phase lag increases with depth, so deeper flaws give weaker signals rotated in phase relative to surface flaws. This phase change is what allows depth estimation.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (phase lag with depth)</p></details></div><div class=\"ptq-question\" id=\"et2-20\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 20.</span> As a differential probe passes over a short, abrupt flaw, what impedance-plane trace is typically produced?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> A single straight line along the lift-off direction</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> No signal, because differential probes cancel all flaws</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> A dot that does not move</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> A figure-eight with two opposing lobes, one per coil</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: D. A figure-eight with two opposing lobes, one per coil</p><p style=\"margin:6px 0;\">Each coil of the pair sees the flaw in turn, so the difference signal swings one way and then the opposite way, tracing two lobes. Only changes seen equally by both coils cancel.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (differential probes)</p></details></div><div class=\"ptq-question\" id=\"et2-21\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 21.</span> A large signal appears as a surface probe approaches the edge of a plate. What is this called, and why does it matter?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Fill-factor effect; it means the coil is too large for the tube</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Edge effect; it can mask flaws close to the edge</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> Skin effect; it shows the test frequency is too low for the plate</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Lift-off; it shows the coating is too thick</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: B. Edge effect; it can mask flaws close to the edge</p><p style=\"margin:6px 0;\">Near an edge the eddy currents cannot complete their normal loops, which produces a large signal unrelated to a flaw. Shielded probes, edge guides or reference scans along the edge help separate edge effect from cracks.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (edge effect)</p></details></div><div class=\"ptq-question\" id=\"et2-22\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 22.</span> For which crack orientation is the eddy current response greatest?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Crack plane parallel to the direction of eddy current flow</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Crack lying parallel to the surface below the coil</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> The response is the same for all orientations</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Crack plane perpendicular to the direction of eddy current flow</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: D. Crack plane perpendicular to the direction of eddy current flow</p><p style=\"margin:6px 0;\">A crack is detected because it interrupts the current path. Maximum interruption, and so maximum response, occurs when the crack lies across the current flow; a crack parallel to the flow causes little disturbance.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (flaw orientation)</p></details></div></section><section class=\"ptq-topic\" id=\"topic-applications-and-techniques\" aria-label=\"Applications and techniques\"><h2>Applications and techniques</h2><div class=\"ptq-question\" id=\"et2-23\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 23.</span> Eddy current lift-off measurement is commonly used to measure what?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> The thickness of paint or anodizing on non-ferrous metals</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> The hardness of ceramic parts after firing</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> The wall thickness of plastic pipe from the outside surface</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> The grain size of austenitic stainless steel welds</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: A. The thickness of paint or anodizing on non-ferrous metals</p><p style=\"margin:6px 0;\">A non-conductive coating holds the probe away from the metal, so the lift-off signal can be calibrated against coating thickness. Ceramics and plastics are not conductive and cannot be tested by eddy currents directly.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET application (coating thickness)</p></details></div><div class=\"ptq-question\" id=\"et2-24\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 24.</span> Compared with a surface-crack inspection, what change is needed to detect corrosion on the far side of an aluminum skin a few millimeters thick?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Use a higher frequency to increase the depth of penetration</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Use a lower frequency so the currents reach the far side</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> Use a smaller lift-off compensation only</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Use a differential probe at the highest frequency available</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: B. Use a lower frequency so the currents reach the far side</p><p style=\"margin:6px 0;\">Penetration increases as frequency decreases (&#x3B4; is proportional to 1/&#x221A;f). Low frequencies are therefore used for far-side and second-layer corrosion, accepting lower sensitivity to small surface cracks.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET practice (frequency selection)</p></details></div><div class=\"ptq-question\" id=\"et2-25\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 25.</span> Why is multi-frequency mixing used in heat-exchanger tube examination?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> To increase the probe travel speed through each tube</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> To measure the hardness of the tube material</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> To suppress support-plate or tubesheet signals so flaws there can be seen</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> To make a bobbin coil respond like a rotating probe to circumferential cracks</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: C. To suppress support-plate or tubesheet signals so flaws there can be seen</p><p style=\"margin:6px 0;\">Signals from different sources change differently with frequency. Combining two or more frequencies lets the instrument cancel a known unwanted signal, such as a support plate, while keeping the flaw response.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET practice (multi-frequency mixing)</p></details></div><div class=\"ptq-question\" id=\"et2-26\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 26.</span> Conductivity measurement on aluminum alloys is commonly used for which purpose?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Detecting laminations in thick plate before machining</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Measuring residual stress directly in welded parts</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> Verifying heat treatment and sorting mixed alloys</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Measuring ultrasonic velocity for thickness gauging</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: C. Verifying heat treatment and sorting mixed alloys</p><p style=\"margin:6px 0;\">Alloy and temper both affect electrical conductivity, so conductivity readings against known limits are used to verify heat treatment and to sort mixed material.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET application (conductivity sorting)</p></details></div></section><section class=\"ptq-topic\" id=\"topic-codes-procedures-and-limitations\" aria-label=\"Codes, procedures and limitations\"><h2>Codes, procedures and limitations</h2><div class=\"ptq-question\" id=\"et2-27\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 27.</span> Why is conventional bobbin-coil eddy current testing of carbon-steel tubing difficult without special techniques?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> High, variable permeability limits penetration and masks flaws</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Carbon steel does not conduct electricity well enough for ET</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> Bobbin coils cannot be made small enough to fit inside steel tubes</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Steel tubes are always too thin-walled to give a usable signal</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: A. High, variable permeability limits penetration and masks flaws</p><p style=\"margin:6px 0;\">Permeability variations in ferromagnetic material give large signals and greatly reduce the standard depth of penetration. Magnetic saturation (as in ASTM E309) or techniques such as remote field testing are used instead.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle; ASTM E309</p></details></div><div class=\"ptq-question\" id=\"et2-28\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 28.</span> Which ASTM standard covers eddy current examination of steel tubular products using magnetic saturation?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> ASTM E426</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> ASTM E243</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> ASTM E309</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> ASTM E2700</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: C. ASTM E309</p><p style=\"margin:6px 0;\">ASTM E309 covers steel tubular products using magnetic saturation. ASTM E426 covers tubular products of austenitic stainless steel, titanium and similar alloys, ASTM E243 covers copper and copper-alloy tubes, and ASTM E2700 covers phased array UT of welds.</p><p style=\"margin:6px 0;color:#475569;\">Basis: ASTM E309</p></details></div><div class=\"ptq-question\" id=\"et2-29\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 29.</span> Which statement is a fundamental limitation of eddy current testing?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> It cannot detect cracks that break the surface of the part</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> It needs a liquid couplant between the probe and the part</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> It cannot be automated for high-volume production testing</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Only conductive materials can be tested, and skin effect limits depth</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: D. Only conductive materials can be tested, and skin effect limits depth</p><p style=\"margin:6px 0;\">Eddy currents need a conductive path and concentrate near the surface, so ET suits surface and near-surface examination of metals. It detects surface cracks well, needs no couplant and is widely automated.</p><p style=\"margin:6px 0;color:#475569;\">Basis: General ET principle (limitations)</p></details></div><div class=\"ptq-question\" id=\"et2-30\" style=\"margin:0 0 22px;padding:14px 16px;border:1px solid #d7e1ea;border-radius:8px;\"><p class=\"ptq-stem\" style=\"margin:0 0 8px;font-weight:600;\"><span class=\"ptq-num\" style=\"color:#075fa8;\">Question 30.</span> During heat-exchanger tube examination, a periodic check on the reference standard shows the system response has changed beyond the procedure tolerance. What is normally required?</p><ol class=\"ptq-options\" type=\"A\" style=\"list-style:none;padding-left:0;margin:0 0 8px;\"><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">A.</span> Continue, and adjust only the results of the remaining tubes</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">B.</span> Restandardize and re-examine tubes done since the last good check</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">C.</span> Average the readings from before and after the check for each tube</li><li style=\"margin:2px 0;\"><span style=\"display:inline-block;min-width:1.6em;font-weight:600;\">D.</span> Record the change and take no further action</li></ol><details class=\"ptq-answer\" style=\"margin:6px 0 0;\"><summary style=\"cursor:pointer;color:#075fa8;font-weight:600;\">Show answer</summary><p style=\"margin:6px 0;\">Correct answer: B. Restandardize and re-examine tubes done since the last good check</p><p style=\"margin:6px 0;\">If the system drifted beyond tolerance, tubes examined since the last good check may have been evaluated wrongly, so the system is restandardized and those tubes are re-examined. ASME Section V Article 8 contains this type of requirement; the procedure sets the tolerance.</p><p style=\"margin:6px 0;color:#475569;\">Basis: ASME BPVC Section V, Article 8</p></details></div></section>\n<section data-course-next-step=\"2026-10\" data-placement=\"end\" aria-label=\"Course next step\" style=\"border-left:4px solid #2563eb;padding:10px 0 10px 16px;margin:24px 0;\"><h2 style=\"font-size:1.25rem;font-weight:700;line-height:1.3;margin:0 0 8px;\">Next step: ET Level 2 course</h2><p style=\"margin:0 0 8px;\">Practice questions check recall. Structured training adds the classroom and hands-on hours your certification route requires. The <a href=\"/et-level-2-training\" style=\"color:#2563eb;text-decoration:underline;\">ET Level 2 course</a> covers impedance-plane analysis, reference standards with machined notches, and surface, bolt-hole and tubing signal interpretation. Completing a course is not certification: under an employer-based scheme such as SNT-TC-1A, your employer certifies you after its own examinations.</p><p style=\"margin:0 0 8px;\"><strong>For employers:</strong> <a href=\"/consulting/ndt-consulting-level-iii\" style=\"color:#2563eb;text-decoration:underline;\">ASNT Level III consulting</a> covers procedures, the written practice, qualification examinations and audit support for your technicians.</p></section>\n<section data-practice-next=\"2026-10\" aria-label=\"Where to go next\" style=\"border-left:4px solid #2563eb;padding:10px 0 10px 16px;margin:24px 0;\"><h2>Where to go next</h2><p>Atlantis NDT courses run year-round, with theory online and hands-on practice on the Practical NDT simulator or in person at the client&apos;s own facility. From here:</p><ul><li><a href=\"/eddy-current-testing-training\">Eddy current testing (ET) training</a>: what Level I and Level II mean in the method and what the practical examination tests.</li><li><a href=\"/practical-ndt\">Practical NDT simulator</a>: browser-based practical training between classroom sessions.</li><li><a href=\"/ndt-practice-tests\">All NDT practice tests</a>: Level II sets for UT, PAUT, TOFD, RT, MT, PT, VT and ET, with exam preparation guides.</li><li><a href=\"/asnt-ndt-level-ii-exam-prep\">ASNT NDT Level II exam preparation</a>: if you are sitting ASNT&apos;s central Level II examination.</li><li><a href=\"/asnt-level-iii-et-exam-prep\">ASNT Level III ET exam preparation</a>: the route once Level II work and experience are behind you.</li></ul><p><a href=\"/contact?service=training&amp;subject=ET%20Level%20II%20training\">Enquire about ET Level II training</a> for yourself or for a team. Tell us the method, the level and where the trainees are based.</p></section>","order":0,"createdAt":"2026-06-24","updatedAt":"2026-10-16T00:00:00Z","h1":"Eddy Current Testing (ET) Level II Practice Questions"}