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Physics Question on Electromagnetism

A square loop PQRS having 10 turns, area 3.6×10−3 m23.6 \times 10^{-3} \, \text{m}^2, and resistance 100 Ω100 \, \Omega is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B=0.5 TB = 0.5 \, \text{T} as shown. Work done in pulling the loop out of the field in 1.0 s1.0 \, \text{s} is ______ ×10−6 J\times 10^{-6} \, \text{J}. loop

Answer

The emf induced in the loop is:
E=NBvℓ,\mathcal{E} = N B v \ell,
where:
v=ℓt.v = \frac{\ell}{t}.
The current induced in the loop is:
i=ER=NBℓ/tR.i = \frac{\mathcal{E}}{R} = \frac{N B \ell / t}{R}.
The force acting is:
F=N⋅i⋅B⋅ℓ=N2B2ℓ2Rt.F = N \cdot i \cdot B \cdot \ell = \frac{N^2 B^2 \ell^2}{R t}.
The work done is:
W=F⋅ℓ=N2B2ℓ2Rt⋅ℓ=N2B2ℓ3Rt.W = F \cdot \ell = \frac{N^2 B^2 \ell^2}{R t} \cdot \ell = \frac{N^2 B^2 \ell^3}{R t}.
Substitute values:
W=(10)2(0.5)2(3.6×10−3)2100⋅1.W = \frac{(10)^2 (0.5)^2 (3.6 \times 10^{-3})^2}{100 \cdot 1}.
W=3.24×10−6 J.W = 3.24 \times 10^{-6} \, \text{J}.
Final Answer: 3.24×10−6 J3.24 \times 10^{-6} \, \text{J}.

Explanation

Solution

The emf induced in the loop is:
E=NBvℓ,\mathcal{E} = N B v \ell,
where:
v=ℓt.v = \frac{\ell}{t}.
The current induced in the loop is:
i=ER=NBℓ/tR.i = \frac{\mathcal{E}}{R} = \frac{N B \ell / t}{R}.
The force acting is:
F=N⋅i⋅B⋅ℓ=N2B2ℓ2Rt.F = N \cdot i \cdot B \cdot \ell = \frac{N^2 B^2 \ell^2}{R t}.
The work done is:
W=F⋅ℓ=N2B2ℓ2Rt⋅ℓ=N2B2ℓ3Rt.W = F \cdot \ell = \frac{N^2 B^2 \ell^2}{R t} \cdot \ell = \frac{N^2 B^2 \ell^3}{R t}.
Substitute values:
W=(10)2(0.5)2(3.6×10−3)2100⋅1.W = \frac{(10)^2 (0.5)^2 (3.6 \times 10^{-3})^2}{100 \cdot 1}.
W=3.24×10−6 J.W = 3.24 \times 10^{-6} \, \text{J}.
Final Answer: 3.24×10−6 J3.24 \times 10^{-6} \, \text{J}.