Question

In figure, a three sided frame if pivoted at AC and hangs vertically, Its sides are each of the same length and have a linear density of 0. 10kg/m. A current of 10.0A is sent through the frame, which is in a uniform magnetic field of 10 mT directed upward. Through what angle will the frame be deflected? A) tan 0 0.05 B) tan 0 0.25 C) tan@ 0.35 D) tan0-1

Answer 1

The dotted line signifies be initial position of the wire magnetic force il B sin theta will be zero for two parallel sides because l B and sin theta = 0 for theta = 0 For the horizontal line F_mag = ilB and the force prints away or but of the page Therefore Torque = ilB times l cos theta = il^2 B cos theta = tau_B The gravitational torque due to three sides lambda = mass per unit length tau_g = lambda lg l sin theta + 2 lambda lg l/2 sin theta = lambda l^2 g sin theta + lambda l^2 g sin theta = 2 lambda l^2 g sin theta for equilibrium these two torques must be equal tau_B = tau_g or, il^2 B cos theta = 2 lambda l^2 g sin theta or, tan theta = iB/2 lambda g or, tan theta = 10 times 10^-3 times 10/2 times 0.1 times 8.8