A square loop (length along one side = 20 cm)rotates in a constant magnetic field which has a magnitude of 2.0 T.At an instant when the angle between the field and the normal to the plane of the loop is equal to 20$\degree$ and increasing at the rate of 10$\degree$/s,what is the magnitude of the induced emf in the loop?
A)13 mV
B)0.27 V
C)4.8 mV
D)14 mV
E)2.2 mV

Question

Quizplus · Accepted Answer

The induced emf in the loop can be found using the formula: 
emf = B*A*(dθ/dt)*cos(θ) 
where B is the magnitude of the magnetic field, A is the area of the loop, (dθ/dt) is the rate of change of the angle between the field and the normal to the plane of the loop, and θ is the angle between the field and the normal to the plane of the loop.

Substituting the given values, we get: 
emf = (2.0 T)*(0.2 m^2)*(10°/s)*cos(20°) 
emf = 0.0048 V = 4.8 mV

Therefore, the best choice is C.

A Square Loop (Length Along One Side = 20 Cm)rotates °\degree°

A Square Loop (Length Along One Side = 20 Cm)rotates $\degree$