Question

A circular coil of radius 5.0 cm and resistance 0.20 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.50e-20t T. What is the magnitude of the current induced in the coil at the time t = 2.0 s?Answera) 2.6 mAb) 7.5 mAc) 9.2 mAd) 4.2 mAe) 1.3 mAShow the calculation

Answer 1

Answer:

(a) 2.6 mA

Explanation:

We are given that

Radius of coil=r=5 cm=$\frac{5}{100}=0.05 m$

1 m=100 cm

Resistance,R=0.20 ohm

Magnetic field,B=$0.50e^{-0.20t} T$

t=2 s

We know that

Emf,$E=-\frac{d(BA)}{dt}=-A\frac{dB}{dt}=-(\pi r^2)\frac{d(0.5e^{-0.20t})}{dt}$

$E=3.14(0.05)^2(0.20)(0.5)e^{-0.20t}$

Substitute t=2

$E=3.14(0.05)^2(0.5)(0.20)e^{-0.20\times 2}=5.26\times 10^{-4} V$

Current,I=$\frac{E}{R}=\frac{5.26\times 10^{-4}}{0.2}=2.6\times 10^{-3} A=2.6 mA$

$1 mA=10^{-3}A$

Answer 2

Answer:

option (A)

Explanation:

radius, r = 5 cm = 0.05 m

resistance, R = 0.20 ohm

Magnetic field, B = 0.5 e ^-0.20t

Let i be the induced current and e is the induced emf

By use of Faraday's law of electromagnetic induction

e = - dФ/dt

where, Ф is the magnetic flux

Ф = B x A

where, B is the strength of magnetic field and A is the area of coil.

$e=- A\frac{dB}{dt}$

$e=- \pi r^{2}\times 0.5 \times (-0.20)e^{-0.20t}$

$e=7.85\times 10^{-4}\times e^{-0.20t}$

induced emf at t = 2 s

$e=7.85\times 10^{-4}\times e^{-0.4}$

e = 5.26 x 10^-4 V

i = e / R = 2.63 x 10^-3 A

i = 2.6 mA

Option (A)