In the figure below, a square closed loop of wire with sides of length 8.4cm lie

Question

In the figure below, a square closed loop of wire with sides of length 8.4cm lies in the (x,y) plane. A non-uniform magnetic field points in the +z-direction (i.e., normal to the square wire loop). (So, the direction of the field is constant; the magnitude is not.)

The magnitude of the field is given by B(t) = (1.7T/(m·s2))t2y. Determine the emf around the square at t = 7.3s. (NOTE: The length of the wire loop is given in centimeters.)

NOTES:

1. Be careful with the units!!!

2. Be sure to enter the correct S.I. units, with the correct S.I. symbol.

Explanation / Answer

Given Data

length of square , y = 8.4cm = 8.4*10^-2 m = 0.084 m

Area of square = y^2 = 0.084^2 = 0.007056 m^2

magnitude of the field B(t) = [1.7 T/(m·s^2)]t^2*y

Therefore dB/dt = d/dt { [1.7 T/(m·s^2)]t^2*y }

            dB/dt = [1.7 T/(m·s^2)] 2t * y

            dB/dt = [3.4]* t * y

t = 7.3s

Solution :-

emf around the square, E = A*dB/dt

                            E = 0.007056 * [3.4]* t * y

                            E = 0.007056 * [3.4]* 7.3 *0.084

                            E = 0.007056 * [3.4]* 7.3 *0.084

                            E = 0.01471 T = 14.71 mT

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