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

Induced current, I = 18.88 A

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

Where

`(d\phi)/(dt)` is the rate of change of magnetic flux,

And
`\phi=BA`

`\epsilon=-NA(dB)/(dt)`

`\epsilon=-78* \pi (0.34)^2((0.4-2.4))/(2)`

`\epsilon=28.32\ V`

Using Ohm's law,
`\epsilon=I* R`

Induced current,
`I=(\epsilon)/(R)`

`I=(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.