# IIT JEE Main / Advanced Physics 2018 – Magnetism: Last Minute Revisioner [ LMR] / Cheat Sheet

Given below is a last minute revisioner of Magnetism for the IIT JEE Main/ Advanced Physics paper . Note that the points covered are based on the exam point of view , and is not meant to be a comprehensive review of the entire magnetism portion.

• A moving charge / flowing electric current produces a magnetic feild ( which is a vector feild ) in addition to the electric force / feild . If a charged particle moves parallel to the magnetic feild , the magnetic force on the charge is zero . The magnetic force differs from an electric force in the fact that a) it depends on the particle velocity b) it’s direction is perpendicular to both |v| and |B| and c) it does no work in displacing a charged particle , provided the magnetic feild is constant.
• Consider a charged particle of mass m moving in an uniform magnetic feild | B | having an initial velocity vector | v | perpendicular to the feild. Then , the path radius = r = mv / qB , angular speed ω = qB/m and time period, T = 2 π m/qB.If the motion in the uniform magnetic feild is at some arbitrary angle θ with respect to B , then the path becomes helical – here there are two velocity components, the perpendicular component moving the charge in a circular path of radius given by r = mv1/qB and the parallel component moving along the feild lines. Pitch p = 2 π m vΦ/qB.
• Consider a charged particle moving in an electric feild which is perpendicular to the magnetic feild . Here , the particle velocity is perpendicular to both the feilds. The magnetic feild will rotate the particle in a circle in the x-z plane , || to the magnetic feild resulting in a helical path with increasing pitch . vx = vo cos ( Bqt/m ) and vz = vo sin ( Bqt/m ) .
• A current carrying conductor of any arbitrary shape in an uniform magnetic feild will experience a force |F| = I|L|x|B| where |L| is the length vector joining initial and final points of a conductor. If it’s a closed loop then  |F| = zero. If the magnetic feild is not uniform , various elements of the loop will experience different forces – means that if the loop is free enough , the loop could have a circular motion , ie , |Fr| = zero, but |ζ | may or may not be zero.
• Considering the force between two infinite , parallel current carrying conductors , the force / unit length = μo I I’/2 π r . Note that the if the direction of current flow is the same , the wires will attract each other.
• As the magnetic feild pattern produced by a small current loop is like a bar magnet, it acts like a magnetic dipole . Here , if the loop is not lying in a single plane , two equal and opposite currents are to be assumed in a single branch [ note that the net change is zero ] and likewise the required loops are to be considered in different planes . For eg , for a cube of side I carrying a current i  , |M| net = – i I² ( iˆ + kˆ ) .
• If a magnetic dipole changes it’s orientation in a magnetic feild by an infinitesimal angular displacement d Φ , the feild does work dW given by ζ d Φ  =  dU , the change in potential energy . The energy present in the loop is U = – |M| |B| .
• A moving point charge with velocity v creates circular magnetic feilds centered on the line of v and which are perpendicular to it. The magnetic feild , |B| = ( μo /4π ) q|v|rˆ/r² .
• For a current carrying straight wire , the feild is zero for points along the length of the wire , but not on it. Here , as the feild is perpendicular to the plane containing both the wire and the point , the lines of force are concentric circles which encircle the wire. For the infinite long wire , if the point is near the end , B = ( μo /4π ) 2I/d and if the point is near one end , then, B = ( μo /4π ) I/d .
• The magnetic feild at the centre of a current carrying arc is B = μo I/2 R b)  a point inside a long solenoid B = μo n I c) for a point at one end of a long solenoid B =  μo n I /2 and d) for a point at a distance R from the centre of a flat strip of width a along it’s perpendicular bisector, B = ( μo i /π a ) tan -¹ ( a/2R )  Note – in cases where the points considered are far from the strip , B = ( μo /2π ) ( i/R ) .

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