Question

A small piece of metal wire is dragged across the gap between the pole pieces of magnet in 10s. The magnetic flux between the pole pieces is 8x10⁻⁴ Wb. Find the magnitude of induced emf.

Answer 1

Given :

Magnetic flux = dΦ = 8x10⁻⁴ Wb

time =dt =10s

Formula to be used:

e= - (dΦ/dt)

=- (8x10⁻⁴) / (10)

=-8x10⁻⁵V

∴magnitude of induced emf is 8x10⁻⁵V

Answer 2

Answer:

Induced EMF, $E=8\times 10^{-5}\ V$

Explanation:

It is given that,

The magnetic flux between the pole pieces is, $\phi=8\times 10^{-4}\ Wb$

The metal wire is dragged across the gap between the pole pieces of magnet in 10 s

We need to find the magnitude of induced emf. Let it is given by E as :

$E=\dfrac{d\phi}{dt}$

$E=\dfrac{8\times 10^{-4}}{10}$

E = 0.00008 V

or

$E=8\times 10^{-5}\ V$

So, the magnitude of induced EMF is $E=8\times 10^{-5}\ V$. Hence, this is the required solution.