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NCERT Exemplar · Q22

Q.A rectangular loop of wire ABCD, of resistance R, lies in the plane of and close to an infinitely long straight wire carrying a current I(t) = I₀(1 − t/T) for 0 ≤ t ≤ T, with I = 0 for t > T (so the current falls linearly from I₀ at t = 0 to zero at t = T). The loop's sides AB and DC, each of length L₁, are parallel to the long wire; the nearer side DC is at perpendicular distance x from the wire and the farther side AB at perpendicular distance x + L₂ (loop breadth L₂). Find the total charge that passes through any given point of the loop during the time interval from t = 0 to t = T.

Rajasthan RbseSubjective· 3mImportance★★★★★
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The total charge equals the change in flux divided by resistance, q=∣Δϕ∣/Rq=|\Delta\phi|/R — the time details of the decay do not matter. The flux through the loop is ϕ=μ0IL12πln⁡x+L2x\phi=\frac{\mu_0 I L_1}{2\pi}\ln\frac{x+L_2}{x}; as the current drops from I0I_0 to 00, q=μ0I0L12πRln⁡x+L2xq=\frac{\mu_0 I_0 L_1}{2\pi R}\ln\frac{x+L_2}{x}.

Concept

The induced current is Iloop=ε/R=−1RdϕdtI_{\text{loop}}=\varepsilon/R=-\dfrac{1}{R}\dfrac{d\phi}{dt}, so the total charge over an interval depends only on the net change of flux:

q=∫Iloop dt=1R∫0T(−dϕdt)dt=∣ϕ(0)−ϕ(T)∣R.q=\int I_{\text{loop}}\,dt=\frac{1}{R}\int_0^T\left(-\frac{d\phi}{dt}\right)dt=\frac{|\phi(0)-\phi(T)|}{R}.

Flux through the loop

With B(x′)=μ0I2πx′B(x')=\dfrac{\mu_0 I}{2\pi x'} and a strip of length L1L_1, width dx′dx':

ϕ=∫xx+L2μ0I2πx′L1 dx′=μ0IL12πln⁡ ⁣(x+L2x).\phi=\int_{x}^{x+L_2}\frac{\mu_0 I}{2\pi x'}L_1\,dx'=\frac{\mu_0 I L_1}{2\pi}\ln\!\left(\frac{x+L_2}{x}\right). …

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