Question

a magnetic flux of 500 μWb per turn,passing through a solenoid of 200 turns, is reversed in direction in 20ms , the average eng induced in the solenoid​

Answer 1

Answer:

See i tried solving this but i am not sure about the answer. Hope you got this as answer.

We know that flux  Ф = NBAcosΘ

so    A =  Ф/NBcosΘ;

as the coil is passing through 180 degrees cos 180 = -1;

putting all this values we get :::

A = - 500 * 10 ^-6 /B

also using Faradey's  Law of Induction

Induces EMF will be  

e = - NdФ/dt = -d(BAcosΘ)/dt

using above equation for area we get

e = -  NB Ф/B(-1)t  

       

( t is dt means time  and -1 is cosΘ which is     negative

and N is number of turns )

e = NФ/t;

e= 200 * 500 * 10^-6 * /20 * 10^-3

e = 5 V.

Explanation:

Answer 2

Answer: 10.0 V

Explanation:

The question provides us with the information that the magnetic flux is reversed. So by that we are expected to use the formula - E (avg.emf) = 2NAB÷t

N= no. of turns in the coil

A= area of cross section

T= (t2 - t1 ) = time

B= magnetic field

=> E =( 2 × 500 × 10^-6 × 200 )

÷ (20 × 10^-3)

=> E = 10.0v

Hope you guys found this useful.