{"id":21139,"date":"2026-01-10T15:24:53","date_gmt":"2026-01-10T07:24:53","guid":{"rendered":"https:\/\/3d-universal.com\/en\/?p=21139"},"modified":"2026-01-10T15:26:54","modified_gmt":"2026-01-10T07:26:54","slug":"magnetism","status":"publish","type":"post","link":"https:\/\/3d-universal.com\/en\/blogs\/magnetism.html","title":{"rendered":"Magnetism: NMAT Physics Review"},"content":{"rendered":"<p><!--more--><\/p>\n<h1>Magnetism: NMAT Physics Review<\/h1>\n<p>Magnetism is a fundamental topic in physics and a recurring subject in the NMAT Physics section. Questions related to magnetic forces, magnetic fields, electromagnetic induction, and the behavior of charged particles in magnetic fields often test both conceptual understanding and problem-solving skills. This comprehensive review is designed to help NMAT examinees build a strong foundation in magnetism, connect formulas with physical meaning, and approach exam questions with confidence.<\/p>\n<h2>Introduction to Magnetism<\/h2>\n<p>Magnetism arises from the motion of electric charges and is one of the four fundamental forces of nature. In everyday life, magnetism is observed in bar magnets, compasses, electric motors, and generators. At the microscopic level, magnetism originates from the orbital motion and spin of electrons inside atoms.<\/p>\n<p>Magnetic effects are closely linked to electricity, and together they form the field of electromagnetism. Understanding magnetism requires familiarity with magnetic poles, magnetic fields, and how magnetic forces interact with moving charges and current-carrying conductors.<\/p>\n<h2>Magnetic Poles and Magnetic Materials<\/h2>\n<p>Every magnet has two poles: a north pole and a south pole. Like poles repel each other, while unlike poles attract. Magnetic poles always exist in pairs; isolating a single magnetic pole (a magnetic monopole) has not been observed in classical physics.<\/p>\n<p>Materials respond differently to magnetic fields and are classified as:<\/p>\n<ul>\n<li><strong>Ferromagnetic materials<\/strong> \u2013 strongly attracted to magnets (e.g., iron, cobalt, nickel)<\/li>\n<li><strong>Paramagnetic materials<\/strong> \u2013 weakly attracted to magnets<\/li>\n<li><strong>Diamagnetic materials<\/strong> \u2013 weakly repelled by magnets<\/li>\n<\/ul>\n<p>Ferromagnetic materials can be permanently magnetized due to the alignment of magnetic domains within the material.<\/p>\n<h2>Magnetic Field and Magnetic Field Lines<\/h2>\n<p>A magnetic field is a region around a magnet or current-carrying conductor where magnetic forces can be detected. It is represented by the symbol <em>B<\/em> and is measured in tesla (T).<\/p>\n<p>Magnetic field lines are imaginary lines used to visualize magnetic fields. They have the following properties:<\/p>\n<ul>\n<li>They emerge from the north pole and enter the south pole outside the magnet<\/li>\n<li>They form closed loops<\/li>\n<li>The closer the lines, the stronger the magnetic field<\/li>\n<li>They never intersect<\/li>\n<\/ul>\n<p>For a bar magnet, the magnetic field is strongest near the poles.<\/p>\n<h2>Magnetic Force on Moving Charges<\/h2>\n<p>A charged particle moving in a magnetic field experiences a magnetic force given by:<\/p>\n<p><strong>F = qvB sin\u03b8<\/strong><\/p>\n<p>where <em>q<\/em> is the charge, <em>v<\/em> is the velocity of the particle, <em>B<\/em> is the magnetic field strength, and \u03b8 is the angle between the velocity vector and the magnetic field.<\/p>\n<p>Important characteristics of magnetic force include:<\/p>\n<ul>\n<li>The force is always perpendicular to both velocity and magnetic field<\/li>\n<li>No work is done by the magnetic force<\/li>\n<li>The speed of the charged particle remains constant<\/li>\n<\/ul>\n<p>This perpendicular force causes charged particles to move in circular or helical paths when entering a magnetic field.<\/p>\n<h2>Right-Hand Rules in Magnetism<\/h2>\n<p>Right-hand rules are essential tools for determining the direction of magnetic forces and fields:<\/p>\n<ul>\n<li><strong>Right-hand rule for force<\/strong>: Point your fingers in the direction of velocity, curl them toward the magnetic field, and your thumb points in the direction of force on a positive charge.<\/li>\n<li><strong>Right-hand thumb rule<\/strong>: Point the thumb in the direction of current; curled fingers show the direction of the magnetic field around a conductor.<\/li>\n<\/ul>\n<p>For negative charges, the force direction is opposite to that given by the right-hand rule.<\/p>\n<h2>Motion of Charged Particles in a Magnetic Field<\/h2>\n<p>When a charged particle enters a magnetic field perpendicular to its velocity, it undergoes uniform circular motion. The radius of the circular path is given by:<\/p>\n<p><strong>r = mv \/ (qB)<\/strong><\/p>\n<p>where <em>m<\/em> is the mass of the particle.<\/p>\n<p>If the velocity has a component parallel to the magnetic field, the particle follows a helical path. This principle is applied in devices such as cyclotrons and mass spectrometers.<\/p>\n<h2>Magnetic Force on a Current-Carrying Conductor<\/h2>\n<p>A current-carrying conductor placed in a magnetic field experiences a force described by:<\/p>\n<p><strong>F = BIL sin\u03b8<\/strong><\/p>\n<p>where <em>I<\/em> is the current, <em>L<\/em> is the length of the conductor in the magnetic field, and \u03b8 is the angle between the conductor and the magnetic field.<\/p>\n<p>This force is the working principle behind electric motors, where electrical energy is converted into mechanical energy.<\/p>\n<h2>Torque on a Current Loop<\/h2>\n<p>A rectangular current loop in a magnetic field experiences a torque that tends to rotate the loop. The torque is given by:<\/p>\n<p><strong>\u03c4 = NIAB sin\u03b8<\/strong><\/p>\n<p>where <em>N<\/em> is the number of turns, <em>A<\/em> is the area of the loop, and \u03b8 is the angle between the normal to the loop and the magnetic field.<\/p>\n<p>This concept is essential for understanding electric motors and galvanometers.<\/p>\n<h2>Magnetic Field Due to Current<\/h2>\n<p>Electric currents produce magnetic fields. Important cases include:<\/p>\n<ul>\n<li><strong>Long straight conductor<\/strong>: B = \u03bc\u2080I \/ (2\u03c0r)<\/li>\n<li><strong>Circular loop<\/strong>: B = \u03bc\u2080I \/ (2R) at the center<\/li>\n<li><strong>Solenoid<\/strong>: B = \u03bc\u2080nI inside the solenoid<\/li>\n<\/ul>\n<p>Here, \u03bc\u2080 is the permeability of free space, and <em>n<\/em> is the number of turns per unit length.<\/p>\n<h2>Electromagnetic Induction<\/h2>\n<p>Electromagnetic induction occurs when a changing magnetic flux induces an electromotive force (emf) in a conductor. This phenomenon is governed by Faraday\u2019s Law:<\/p>\n<p><strong>\u03b5 = &#8211; d\u03a6 \/ dt<\/strong><\/p>\n<p>The negative sign represents Lenz\u2019s Law, which states that the induced emf opposes the change in magnetic flux that produced it.<\/p>\n<p>Magnetic flux is defined as:<\/p>\n<p><strong>\u03a6 = BA cos\u03b8<\/strong><\/p>\n<h2>Lenz\u2019s Law and Conservation of Energy<\/h2>\n<p>Lenz\u2019s Law ensures the conservation of energy in electromagnetic processes. If the induced current enhanced the change in magnetic flux instead of opposing it, energy would be created from nothing.<\/p>\n<p>NMAT questions often test conceptual understanding of Lenz\u2019s Law by asking for the direction of induced current when a magnet approaches or recedes from a loop.<\/p>\n<h2>Applications of Magnetism<\/h2>\n<p>Magnetism plays a crucial role in modern technology, including:<\/p>\n<ul>\n<li>Electric motors and generators<\/li>\n<li>Transformers<\/li>\n<li>Magnetic resonance imaging (MRI)<\/li>\n<li>Hard drives and data storage<\/li>\n<li>Speakers and microphones<\/li>\n<\/ul>\n<p>Understanding the underlying physics helps NMAT examinees connect theoretical concepts to real-world applications.<\/p>\n<h2>Common NMAT Pitfalls in Magnetism<\/h2>\n<p>Students often make mistakes in magnetism-related questions due to:<\/p>\n<ul>\n<li>Incorrect application of right-hand rules<\/li>\n<li>Forgetting that magnetic force does no work<\/li>\n<li>Confusing electric and magnetic fields<\/li>\n<li>Ignoring angle dependence in formulas<\/li>\n<\/ul>\n<p>Careful reading of the question and proper vector analysis can help avoid these errors.<\/p>\n<h2>NMAT Exam Tips for Magnetism<\/h2>\n<p>To excel in magnetism questions on the NMAT:<\/p>\n<ul>\n<li>Memorize key formulas and understand their derivations<\/li>\n<li>Practice direction-based problems using right-hand rules<\/li>\n<li>Visualize magnetic fields and forces<\/li>\n<li>Review common applications like motors and generators<\/li>\n<\/ul>\n<p>Conceptual clarity combined with regular practice is the key to mastering magnetism for the NMAT.<\/p>\n<h2>Conclusion<\/h2>\n<p>Magnetism is a vital component of the NMAT Physics syllabus and an extension of fundamental concepts in electricity and motion. By understanding magnetic fields, forces on charges and currents, and electromagnetic induction, students can confidently tackle a wide range of exam questions. This review serves as a solid foundation, but consistent problem-solving and conceptual reinforcement are essential for achieving a high NMAT score.<\/p>\n<h2>Problem Sets<\/h2>\n<ol>\n<li>A proton (charge +1.6 \u00d7 10<sup>-19<\/sup> C) moves with a speed of 3.0 \u00d7 10<sup>6<\/sup> m\/s perpendicular to a uniform magnetic field of 0.50 T.<br \/>\nWhat is the magnitude of the magnetic force acting on the proton?<\/li>\n<li>An electron enters a uniform magnetic field of 0.20 T with a velocity of 4.0 \u00d7 10<sup>6<\/sup> m\/s at an angle of 30\u00b0 to the field.<br \/>\nWhat is the magnitude of the magnetic force on the electron? (Use |q| = 1.6 \u00d7 10<sup>-19<\/sup> C)<\/li>\n<li>A charged particle with charge +2.0 \u00d7 10<sup>-6<\/sup> C moves at 500 m\/s through a magnetic field of 0.80 T.<br \/>\nIf the force measured is 6.4 \u00d7 10<sup>-4<\/sup> N, what is the angle between the velocity and the magnetic field?<\/li>\n<li>A wire of length 0.40 m carries a current of 8.0 A in a magnetic field of 0.60 T.<br \/>\nIf the wire is perpendicular to the magnetic field, what is the force on the wire?<\/li>\n<li>A straight wire carrying 12 A is placed in a 0.50 T magnetic field. The wire length in the field is 0.25 m and makes an angle of 60\u00b0 with the field.<br \/>\nFind the magnitude of the force on the wire.<\/li>\n<li>A proton moves in a circle of radius 0.12 m in a uniform magnetic field of 0.80 T (perpendicular entry).<br \/>\nIf the proton mass is 1.67 \u00d7 10<sup>-27<\/sup> kg and charge is 1.6 \u00d7 10<sup>-19<\/sup> C, what is the proton\u2019s speed?<\/li>\n<li>An electron moves perpendicular to a magnetic field of 0.10 T and follows a circular path of radius 2.0 cm.<br \/>\nIf the electron mass is 9.11 \u00d7 10<sup>-31<\/sup> kg and charge magnitude is 1.6 \u00d7 10<sup>-19<\/sup> C, find the electron\u2019s speed.<\/li>\n<li>A long straight wire carries a current of 5.0 A.<br \/>\nWhat is the magnitude of the magnetic field at a point 4.0 cm from the wire? (Use \u03bc<sub>0<\/sub> = 4\u03c0 \u00d7 10<sup>-7<\/sup> T\u00b7m\/A)<\/li>\n<li>A circular loop of radius 0.10 m carries a current of 3.0 A.<br \/>\nWhat is the magnetic field at the center of the loop? (Use \u03bc<sub>0<\/sub> = 4\u03c0 \u00d7 10<sup>-7<\/sup> T\u00b7m\/A)<\/li>\n<li>A solenoid has 1200 turns and a length of 0.60 m, carrying a current of 2.5 A.<br \/>\nFind the magnetic field inside the solenoid. (Use \u03bc<sub>0<\/sub> = 4\u03c0 \u00d7 10<sup>-7<\/sup> T\u00b7m\/A)<\/li>\n<li>A coil has 50 turns and area 0.020 m<sup>2<\/sup>. The magnetic field through the coil changes uniformly from 0.10 T to 0.60 T in 0.20 s, with the field perpendicular to the coil.<br \/>\nWhat is the magnitude of the induced emf?<\/li>\n<li>A circular loop of area 0.015 m<sup>2<\/sup> is in a uniform magnetic field of 0.40 T. The loop is rotated so that the angle between the field and the loop\u2019s normal changes from 0\u00b0 to 60\u00b0 in 0.50 s.<br \/>\nWhat is the average induced emf? (Assume 1 turn.)<\/li>\n<li>A conducting rod of length 0.50 m moves at 6.0 m\/s perpendicular to a magnetic field of 0.30 T.<br \/>\nWhat is the motional emf induced across the rod?<\/li>\n<li>A rectangular loop is pulled out of a uniform magnetic field. As it leaves the field region, the magnetic flux through it decreases.<br \/>\nAccording to Lenz\u2019s law, does the induced current create a magnetic field in the same direction as the original field or the opposite direction?<\/li>\n<li>A particle with charge +3.2 \u00d7 10<sup>-19<\/sup> C enters a magnetic field of 0.40 T with velocity 2.0 \u00d7 10<sup>6<\/sup> m\/s perpendicular to the field.<br \/>\nWhat is the period of its circular motion if its mass is 6.6 \u00d7 10<sup>-27<\/sup> kg?<\/li>\n<\/ol>\n<h2>Answer Keys<\/h2>\n<ol>\n<li><strong>F = qvB sin\u03b8<\/strong>, \u03b8 = 90\u00b0 \u2192 sin\u03b8 = 1<br \/>\nF = (1.6 \u00d7 10<sup>-19<\/sup>)(3.0 \u00d7 10<sup>6<\/sup>)(0.50)<br \/>\nF = 2.4 \u00d7 10<sup>-13<\/sup> N<\/li>\n<li>F = |q|vB sin30\u00b0 = (1.6 \u00d7 10<sup>-19<\/sup>)(4.0 \u00d7 10<sup>6<\/sup>)(0.20)(0.5)<br \/>\nF = 6.4 \u00d7 10<sup>-14<\/sup> N<\/li>\n<li>F = qvB sin\u03b8 \u2192 sin\u03b8 = F\/(qvB)<br \/>\nsin\u03b8 = (6.4 \u00d7 10<sup>-4<\/sup>)\/[(2.0 \u00d7 10<sup>-6<\/sup>)(500)(0.80)]<br \/>\nDenominator = 2.0 \u00d7 10<sup>-6<\/sup> \u00d7 500 = 1.0 \u00d7 10<sup>-3<\/sup>; then \u00d7 0.80 = 8.0 \u00d7 10<sup>-4<\/sup><br \/>\nsin\u03b8 = (6.4 \u00d7 10<sup>-4<\/sup>)\/(8.0 \u00d7 10<sup>-4<\/sup>) = 0.80<br \/>\n\u03b8 \u2248 53\u00b0<\/li>\n<li>F = BIL sin90\u00b0 = (0.60)(8.0)(0.40) = 1.92 N<\/li>\n<li>F = BIL sin60\u00b0 = (0.50)(12)(0.25)(0.866)<br \/>\n(0.50)(12) = 6; 6(0.25) = 1.5; 1.5(0.866) \u2248 1.30 N<\/li>\n<li>r = mv\/(qB) \u2192 v = r(qB)\/m<br \/>\nv = (0.12)(1.6 \u00d7 10<sup>-19<\/sup>)(0.80)\/(1.67 \u00d7 10<sup>-27<\/sup>)<br \/>\nNumerator = 0.12 \u00d7 1.28 \u00d7 10<sup>-19<\/sup> = 1.536 \u00d7 10<sup>-20<\/sup><br \/>\nv \u2248 (1.536 \u00d7 10<sup>-20<\/sup>)\/(1.67 \u00d7 10<sup>-27<\/sup>) \u2248 9.2 \u00d7 10<sup>6<\/sup> m\/s<\/li>\n<li>v = r(qB)\/m<br \/>\nv = (0.020)(1.6 \u00d7 10<sup>-19<\/sup>)(0.10)\/(9.11 \u00d7 10<sup>-31<\/sup>)<br \/>\nNumerator = 3.2 \u00d7 10<sup>-22<\/sup><br \/>\nv \u2248 (3.2 \u00d7 10<sup>-22<\/sup>)\/(9.11 \u00d7 10<sup>-31<\/sup>) \u2248 3.5 \u00d7 10<sup>8<\/sup> m\/s<\/li>\n<li>B = \u03bc<sub>0<\/sub>I\/(2\u03c0r)<br \/>\nB = (4\u03c0 \u00d7 10<sup>-7<\/sup>)(5.0)\/(2\u03c0 \u00d7 0.040)<br \/>\nCancel \u03c0: B = (4 \u00d7 10<sup>-7<\/sup> \u00d7 5.0)\/(2 \u00d7 0.040)<br \/>\nNumerator = 2.0 \u00d7 10<sup>-6<\/sup>; Denominator = 0.080<br \/>\nB = 2.5 \u00d7 10<sup>-5<\/sup> T<\/li>\n<li>B = \u03bc<sub>0<\/sub>I\/(2R)<br \/>\nB = (4\u03c0 \u00d7 10<sup>-7<\/sup>)(3.0)\/(2 \u00d7 0.10)<br \/>\nB = (12\u03c0 \u00d7 10<sup>-7<\/sup>)\/0.20 = 60\u03c0 \u00d7 10<sup>-7<\/sup><br \/>\nB \u2248 1.9 \u00d7 10<sup>-5<\/sup> T<\/li>\n<li>B = \u03bc<sub>0<\/sub>nI, where n = N\/L = 1200\/0.60 = 2000 m<sup>-1<\/sup><br \/>\nB = (4\u03c0 \u00d7 10<sup>-7<\/sup>)(2000)(2.5)<br \/>\n2000 \u00d7 2.5 = 5000<br \/>\nB = 4\u03c0 \u00d7 10<sup>-7<\/sup> \u00d7 5000 = 2\u03c0 \u00d7 10<sup>-3<\/sup><br \/>\nB \u2248 6.3 \u00d7 10<sup>-3<\/sup> T<\/li>\n<li>\u03b5 = N A (\u0394B\/\u0394t) (since perpendicular, cos\u03b8 = 1)<br \/>\n\u03b5 = 50(0.020)[(0.60 &#8211; 0.10)\/0.20]<br \/>\n\u0394B\/\u0394t = 0.50\/0.20 = 2.5 T\/s<br \/>\n\u03b5 = 50(0.020)(2.5) = 1.0(2.5) = 2.5 V<\/li>\n<li>Average emf: \u03b5<sub>avg<\/sub> = |\u0394\u03a6|\/\u0394t, \u03a6 = BA cos\u03b8<br \/>\nInitial: \u03b8 = 0\u00b0 \u2192 \u03a6<sub>i<\/sub> = BA(1) = (0.40)(0.015) = 0.006 Wb<br \/>\nFinal: \u03b8 = 60\u00b0 \u2192 \u03a6<sub>f<\/sub> = BA cos60\u00b0 = 0.006(0.5) = 0.003 Wb<br \/>\n|\u0394\u03a6| = 0.003 Wb<br \/>\n\u03b5<sub>avg<\/sub> = 0.003\/0.50 = 0.006 V<\/li>\n<li>Motional emf: \u03b5 = BLv<br \/>\n\u03b5 = (0.30)(0.50)(6.0) = 0.90 V<\/li>\n<li>The loop opposes the decrease in flux, so it produces a magnetic field in the <strong>same direction<\/strong> as the original field.<\/li>\n<li>Period in magnetic field: T = 2\u03c0m\/(qB)<br \/>\nT = 2\u03c0(6.6 \u00d7 10<sup>-27<\/sup>)\/[(3.2 \u00d7 10<sup>-19<\/sup>)(0.40)]<br \/>\nDenominator = 1.28 \u00d7 10<sup>-19<\/sup><br \/>\nT \u2248 2\u03c0(5.16 \u00d7 10<sup>-8<\/sup>) \u2248 3.24 \u00d7 10<sup>-7<\/sup> s<\/li>\n<\/ol>\n<blockquote class=\"wp-embedded-content\" data-secret=\"HXFM5p334W\"><p><a href=\"https:\/\/3d-universal.com\/en\/blogs\/nmat-physics-review.html\">NMAT Physics Review: NMAT Study Guide<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; visibility: hidden;\" title=\"&#8220;NMAT Physics Review: NMAT Study Guide&#8221; &#8212; Study English at 3D ACADEMY, a Language School in Cebu, Philippines\" src=\"https:\/\/3d-universal.com\/en\/blogs\/nmat-physics-review.html\/embed#?secret=K2nYn1X0P7#?secret=HXFM5p334W\" data-secret=\"HXFM5p334W\" width=\"500\" height=\"282\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n<blockquote class=\"wp-embedded-content\" 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