A 6 cm x 8 cm rectangular coil in the xz-plane has 50 turns of wire carrying cur

Question

A 6 cm x 8 cm rectangular coil in the xz-plane has 50 turns of wire carrying current 2 A, (a) Find the magnitude and direction of the magnetic dipole moment [mu = 0.48 A.m^2, -y]. (b) A uniform 1.5-T external magnetic field is applied in the +x-direction. (i) Calculate the force on each side of the coil, and (ii) from this find the net torque [0.72 N.m]. (iii) Show agreement with the formula tau mu B. (iv) Evaluate tau when the coil has rotated through 53 degree. (v) In which plane will the coil lie when in equilibrium-when tau 0?

Explanation / Answer

Given,

rectangular coil of length = 6 cm or 6 * 10^-2 m

and breadth = 8 * 10^-2 m

number of turns = 50

current = 2 A

magnetic diapole moment = current * area * number of turns

magnetic diapole moment = 2 * 6 * 10^-2 * 8 * 10^-2 * 50

magnetic diapole moment = 0.48 A.m^2

direction of magnetic diapole moment is perpendicular to the current loop in the right-hand-rule direction that is -y

magnetic field = 1.5 T

force on 6 cm sides will be 0 since they are in the direction of magnetic field

force on 8 cm side = current * length * magnetic field * number of turns

force on 8 cm side = 2 * 8 * 10^-2 * 1.5

force on 8 cm side = 12 N

torque = force * distance

torque = 12 * 3 * 10^-2

torque = 0.36 Nm

since 2 sides are there so net torque = 2 * 0.36

net torque = 0.72 Nm

also torque = magnetic moment * magnetic field

torque = 0.48 * 1.5

torque = 0.72 Nm

so this is with agreement with torque = magnetic moment * magnetic field

when coil is rotated through 53 degree then

torque = magnetic moment * magnetic field * sin (53)

torque = 0.72 * sin(53)

torque when coil rotated through 53 degree= 0.575 Nm

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