The external magnetic field in a loop decreases from 0.1 Tesla to zero in 2 seconds. The loop of wire has ten turns, a radius of 10 cm, with its axis 45 degrees relative to the magnetic field. How large is the induced voltage in the loop?
A. 0 mV
B. 11 mV
C. 9.4 mV
D. 32.5 mV
E. 0.31 mV

Question

The external magnetic field in a loop decreases from 0.1 Tesla to zero in 2 seconds. The loop of wire has ten turns, a radius of 10 cm, with its axis 45 degrees relative to the magnetic field. How large is the induced voltage in the loop?
	A. 0 mV
	B. 11 mV
	C. 9.4 mV
	D. 32.5 mV
	E. 0.31 mV

anonymous · Answer

The induced voltage is given as "rate of change of magnetic flux".
Flux is given as $$\Phi = NAB _{p}$$

N = no of turns in the coil
A= area of the coil
"Bp" = magnitude of magnetic field perpendicular to the area 
Initially Bp = 0.1 cos 45
Finally Bp = 0

Therefore, initial flux = N*A* 0.1*cos(45)
                 final flux = 0
So change in flux = N*A*0.1*cos(45) - 0   = N*A*0.1*cos(45)
Time taken for this change to happen = 2seconds.

Voltage =change in flux/time taken
          =N*A*0.1*cos(45) / 2

Put the value of N and A and find the voltage.