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

Answer:

Induced current, I = 0.5 A

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 :

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

Where

$\phi=BA$

$\epsilon=\dfrac{-d(NBA)}{dt}$

$\epsilon=NA\dfrac{B_f-B_i}{t}$

$\epsilon=20\times 0.005\times \dfrac{6-2}{2}$

$\epsilon=0.2\ V$

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

$\epsilon=I\times R$

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{0.2}{0.4}$

I = 0.5 A

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