A wire loop with 10 turns, as shown in the figure, lines in a horizontal plane.

Question

A wire loop with 10 turns, as shown in the figure, lines in a horizontal plane. It is a rectangle 10.0 cm by 5.0 cm (see figure). It is parallel to a uniform magnetic field that is also horizontal (arrows in figure). The loop carries a current of 2.0 A. This loop can freely rotate about a nonmagnetic axle through its center. On the right edge of the loop, a 50 g mass is hanging from a massless thread. If we want to prevent the loop from rotating around the axle, what is the magnitude of the magnetic field?

Explanation / Answer

force is NOT generated in a wire if it lies in the same direction as magnetic field. Thus, in this case, there will be a force only across the 10cm portions of the wire loop and NOT the 5cm ones. The force will be equal to nBIl where n is number of turns. Thus, force on either side will be 10 x B x 2.0A x 0.1m = 2B where B is strength of magnetic field. Assume that force on the left hand side of the loop is downwards and on the right (where the 50g mass lies) is upwards. Moments across the axle due to these forces will be 2B x 2.5cm = 0.025 x 2B = 0.05B EACH. so total moments = 0.1B. This must be balanced out by the torque exerted my the 50g mass. this torque = weight x perpendicular distance = 0.05Kg x 9.8 x 0.025m = 0.01225. Thus, for equilibrium, 0.1B = 0.01225 and hence, B = 0.1225

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