A 5-turn square loop (10 cm along a side, resistance = 4.0 Ω) is placed in a magnetic field that makes an angle of 30° with the plane of the loop. The magnitude of this field varies with time according to B = 0.50t2, where t is measured in s and B in T. What is the induced current in the coil at t = 4.0 s?
A)25 mA
B)5.0 mA
C)13 mA
D)43 mA
E)50 mA

Question

Quizplus · Accepted Answer

Explanation:The magnetic flux through the loop is given by:$$\Phi = BA\cos	heta$$where B is the magnetic field, A is the area of the loop, and $	heta$ is the angle between the magnetic field and the normal to the loop.The induced emf is given by:$$\varepsilon = -\frac{d\Phi}{dt}$$Substituting the given values, we get:$$\varepsilon = -5(0.50t^2)(0.10)^2\cos30^\circ = -0.0125t^2$$The induced current is given by:$$I = \frac{\varepsilon}{R} = \frac{-0.0125t^2}{4.0} = -0.003125t^2$$At t = 4.0 s, the induced current is:$$I = -0.003125(4.0)^2 = -0.050A = -50mA$$The negative sign indicates that the current is flowing in the opposite direction to the assumed direction.

A 5-Turn Square Loop (10 Cm Along a Side, Resistance