A square loop of area 2-5-10-3 m2 and having 100 turns with a total resistance of 100- is moved out of a uniform magnetic field of 0-40T in 1s with a constant speed- Then the work done in pulling the loop is -

The correct option is D 0.1mJ Area = 2.5 × 10^ 3 Side of square = a = 0.05 m #160; #160;The frame is pulled out in 1 sVelocity = dxdt ...

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Byju's Answer

Standard XII

Physics

Magnetic Flux and Faraday's Law

A square loop...

Question

A square loop of area $2.5 \times 10^{- 3}$ $m^{2}$ and having $100$ turns with a total resistance of $100 Ω$ is moved out of a uniform magnetic field of $0.40 T$ in $1 s$ with a constant speed. Then the work done in pulling the loop is :

0

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1 m J

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1 μ J

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0.1 m J

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Solution

The correct option is D $0.1 m J$
Area = $2.5 \times 10^{- 3}$
Side of square = $a = 0.05 m$
The frame is pulled out in 1 s
Velocity = $\frac{d x}{d t} = \frac{a}{1 s} = 0.05 m / s$
By faraday's law
$ϵ = - \frac{d ϕ}{d t} = - \frac{d N (2.5 \times 10^{- 3}) 0.4}{d t} = 0.1 V$
Resistance = $100$ $Ω$
Work done = $\frac{ϵ^{2}}{R} = 0.1 m J$

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A square loop of area 2.5×10−3 m2 and having 100 turns with a total resistance of 100Ω is moved out of a uniform magnetic field of 0.40T in 1s with a constant speed. Then the work done in pulling the loop is :

The correct option is D 0.1mJArea = 2.5×10−3Side of square = a=0.05m The frame is pulled out in 1 sVelocity = dxdt=a1s=0.05m/sBy faraday's lawϵ=−dϕdt=−dN(2.5×10−3)0.4dt=0.1VResistance = 100 ΩWork done = ϵ2R=0.1mJ

A square loop of area $2.5 \times 10^{- 3}$ $m^{2}$ and having $100$ turns with a total resistance of $100 Ω$ is moved out of a uniform magnetic field of $0.40 T$ in $1 s$ with a constant speed. Then the work done in pulling the loop is :