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

option (A)

Step-by-step 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(dB)/(dt)`

`e=- \pi r^(2)* 0.5 * (-0.20)e^(-0.20t)`

`e=7.85* 10^(-4)* e^(-0.20t)`

induced emf at t = 2 s

`e=7.85* 10^(-4)* 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)

Answer 2

(a) 2.6 mA

Step-by-step explanation:

We are given that

Radius of coil=r=5 cm=
`(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=-(d(BA))/(dt)=-A(dB)/(dt)=-(\pi r^2)(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* 2)=5.26* 10^(-4) V`

Current,I=
`(E)/(R)=(5.26* 10^(-4))/(0.2)=2.6* 10^(-3) A=2.6 mA`

`1 mA=10^(-3)A`