Answer to Question #113263 in Electricity and Magnetism for nurfarhanah

Question

A current carrying rectangular wire loop with width a = 0.12m and length b= 0.2m is in the xy plane, supported by a non-conducting , frictionless axle of negligible weight. A current of I = 3A travels counterclockwise in the circuit. Calculate the magnitude and direction of the force exerted on a) left, b) right segments of wire by a uniform magnetic field of 0.250 T that points +x-direction, c) magnetic force exerted on the top, d) magnetic force exerted on the bottom segments and e) find the magnitude of net torque on the loop about the axle.

Answer 1

force due to magnetic field is i($l\times B$ )

as the B is in positive x then the magnetic force will be generated in the y direction but not in the x direction

substituting a or b in place of l in the above formula we can find the value of force but one thing which we must remember is that the force will be exerted on that part of the coil which lies in the direction parallel to the x axis

torque is calculated by $\tau=\mu\times B$

where $\mu=IA$ where I is the current and A is the area

$\mu=0.024\times 3=0.072$ in the -z direction

$\tau=0.072\times0.25=0.018$