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. T

Question

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. T

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

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - 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`