Question

The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from 2.40 To 0.400 T in 2.00 s. If the resistance of the coil is 1.50 Ω, what is the magnitude of the induced current in the coil

Answer 1

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Where

$\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux,

And $\phi=BA$

$\epsilon=-NA\dfrac{dB}{dt}$

$\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}$

$\epsilon=28.32\ V$

Using Ohm's law, $\epsilon=I\times R$

Induced current, $I=\dfrac{\epsilon}{R}$

$I=\dfrac{28.32}{1.5}$

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.