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A square loop of wire with side length 1...

A square loop of wire with side length 10 cm is placed at angle of `45^(@)` with a magnetic field that changes uniformly from 0.1 T to zero in 0.7s. The induced current in the loop (its resistance is `1Omega`) is

A

1.0 mA

B

2.5 mA

C

3.5 mA

D

4.0 mA

Text Solution

Verified by Experts

The correct Answer is:
A

Area of square loop, S = 10 `cm xx10 cm`
`rArrS=100cm^(2)=100xx10^(-4)m^(2)=10^(-2)m^(2)`
Initial magnetic flux linked with loop,
`phi_(B_(1))=B_(1)Scosphi=0.1xx10^(-2)xxcos45^(@)`
`=(0.1xx10^(-2)xx1)/(sqrt2)=(10^(-3))/(sqrt2)Wb`
Final magnetic flux linked with loop,
`phi_(B_(2))=0Wb" "[thereforeB_(0)=0]`
The induced emf in the loop, `epsilon=-(dphi)/(dt)=-((phi_(2)-phi_(1)))/(t)`
`=((0-(10^(-3))/(sqrt2)))/(0.7)=(10^(-3))/(0.7xxsqrt2)~=10^(-3)V`
The induced current in the loop,
`I=(epsilon)/(R)=(10^(-3)V)/(1Omega)=10^(-3)A=1.0mA`
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