Question

Suppose the first coil of wire illustrated in the simulation is wound around the iron core 7 times. Suppose the second coil is wound around the core N2 = 13 times. Shortly after the switch has been closed, the flux through the second coil of wire changes from 2.1 Wb per turn to 2.7 Wb in each turn of wire over a time interval of 0.3 s. Find the magnitude of the emf induced in the coil.

Answer 1

Answer:

Emf induced in the coli will be equal to 26 volt

Explanation:

We have given number of turns N = 13

It is given that magnetic flux changes from 2.1 Weber to 2.7 Weber in 0.3 sec

Change in flux $d\Phi =2.7-2.1=0.6Weber$

Change in time dt = 0.3 sec

We have to find the induced emf

Induced emf is equal to $e=N\frac{d\Phi }{dt}=13\times \frac{0.6}{0.3}=26volt$

So emf induced in the coil will be equal to 6 volt

Answer 2

Answer:

$emf=26\ V$

Explanation:

Given:

no. of turns in the first coil, $n_1=7$

no. of turns in the second coil, $n_2=13$

the change in flux through the secondary coil, $d\phi=2.7-2.1=0.6\ Wb$

time taken for the change in flux, $dt=0.3\ s$

• According to the Faraday's law there will be an emf induced in the coil associated with the rate of change in magnetic flux.

This emf is mathematically given as:

$emf=n_2\times \frac{d\phi}{dt}$

$emf=13\times \frac{0.6}{0.3}$ (we take no. of turns of the second coil because the rate of change in flux is associated with the second coil)

$emf=26\ V$