Square loop in Magnetic Field. Assume a square loop of total mass 140 g uniforml

Question

Square loop in Magnetic Field. Assume a square loop of total mass 140 g uniformly distributed along its perimeter. A current of magnitude 2.0 A is running in the loop and the loop is held in a uniform magnetic field of strength 3.00 T applied so that its plane is parallel to the field. (In particular, the field is parallel to two sides of the loop and perpendicular to the other two sides.) Recalling that the moment of inertia I_zod (not to be confused with current) of a rod of mass M and length l about its center of mass is 1/12 M_rodl^2, find the initial angular acceleration of the loop once it is released.

Explanation / Answer

Torque in a square loop in a parallal magneteic field = IbaB
=> 2**3*l^2 = Im * alpha
Im = moment of inertia of the loop about the rotaional axis = ml^2 / 6
alpha = 36/m = 3600/14 = 257.143 rad/s^2

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