Question

A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle of 60o with the plane of the coil. If the magntidue of this field increases by 6.0 mT every 10 ms, what is the magnitude of the emf induced in the coil?a)60 mVb)46 mVc)50 mVd)55 mv

Answer 1

Answer:

The magnitude of the emf induced in the coil is 60 mV.

Explanation:

We have,

Side of the square coil, a = 24 cm = 0.24 m

Number of turns in the coil, N = 2

It is placed in a uniform magnetic field that makes an angle of 60 degrees with the plane of the coil. If the magnitude of this field increases by 6.0 mT every 10 ms, we need to find the magnitude of the emf induced in the coil.

We know that the induced emf is given by the rate of change of magnetic flux throughout the coil. So,

$\epsilon=N\dfrac{d\phi}{dt}\\\\\epsilon=N\dfrac{d(BA\cos \theta)}{dt}$

$\theta$ is the angle between magnetic field and the normal to area vector.

But in this case, a uniform magnetic field that makes an angle of 60 degrees with the plane of the coil. So, the angle between magnetic field and the normal to area vector is 90-60=30 degrees.

Now, induced emf becomes :

$\epsilon=N\dfrac{d(BA\cos \theta)}{dt}\\\\\epsilon=NA\dfrac{dB}{dt}\cos \theta\\\\\epsilon=2\times (0.24)^2\times \dfrac{6\times 10^{-3}}{10\times 10^{-3}}\times \cos (30)\\\\\epsilon=0.0598\ V\\\\\epsilon=59.8\ V$

or

$\epsilon=60\ mV$

So, the magnitude of the emf induced in the coil is 60 mV. Hence, this is the required solution.