Question

A 200 turn coil is in a uniform magnetic field that is decreasing at the rate 0.20 T/s. The coil is perpendicular to the field and its dimensions are 0.20 m by 0.40 m. What is the magnitude of the induced emf in the coil

Answer 1

Answer:

emf = 3.2V

Explanation:

In order to calculate the magnitude of the induced emf in the coil you use the following formula:

$emf=-N\frac{d\Phi_B}{dt}$       (1)

N: turns of the coil = 200

ФB: magnetic flux = A*B

A: area of the coil = (0.20m)(0.40m) = 0.08m²

B: magnitude of the magnetic field

You take into account that the area of the coil is constant, while magnetic field changes on time. Then, the equation (1) becomes:

$emf=-NA\frac{dB}{dt}$         (2)

dB/dt =  rate of change of the magnetic field = -0.20T/s (it is decreasing)

You replace the values of all parameters in the equation (2):

$emf=-200(0.08m^2)(-0.20T/s)=3.2V$

The induced emf in the coil is 3.2V