Question

A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40 Ω, what is the magnitude of the induced current?

a. 0.70 A
b. 0.60 A
c. 0.50 A
d. 0.80 A
e. 0.20 A

Answer 1

Induced current, I = 0.5 A

Step-by-step explanation:

It is given that,

number of turns, N = 20

Area of wire,
`A=50\ cm^2=0.005\ m^2`

Initial magnetic field,
`B_i=2\ T`

Final magnetic field,
`B_f=6\ T`

Time taken, t = 2 s

Resistance of the coil, R = 0.4 ohms

We know that due to change in magnetic field and emf will be induced in the coil. Its formula is given by :

`\epsilon=(-d\phi)/(dt)`

Where

`\phi=BA`

`\epsilon=(-d(NBA))/(dt)`

`\epsilon=NA(B_f-B_i)/(t)`

`\epsilon=20* 0.005* (6-2)/(2)`

`\epsilon=0.2\ V`

Let I is the induced current in the wire. It can be calculated using Ohm's law as :

`\epsilon=I* R`

`I=(\epsilon)/(R)`

`I=(0.2)/(0.4)`

I = 0.5 A

So, the magnitude of the induced current in the coil is 0.5 A. Hence, this is the required solution.