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

`emf=26\ V`

Step-by-step 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* (d\phi)/(dt)`

`emf=13* (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`

Answer 2

Emf induced in the coli will be equal to 26 volt

Step-by-step 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(d\Phi )/(dt)=13* (0.6)/(0.3)=26volt`

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