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NCERT Exemplar · Q21

Q.An infinitely long straight wire lies in the plane of the paper and carries a current I(t) whose rate of change dI/dt = λ is constant. A rectangular loop ABCD of wire, of resistance R, lies in the same plane with its two sides AB and DC (each of length l) parallel to the long wire. The nearer side DC is at a perpendicular distance x₀ from the wire and the farther side AB at a perpendicular distance x₀ + r (so the loop's breadth measured away from the wire is r). Find the magnitude of the current induced in the loop.

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The long wire's field falls off as 1/x1/x, so the flux through the loop is found by integrating across its breadth: ϕ=μ0Il2πln⁡x0+rx0\phi=\dfrac{\mu_0 Il}{2\pi}\ln\dfrac{x_0+r}{x_0}. Since dI/dt=λdI/dt=\lambda is constant, the induced emf is constant and Iloop=μ0lλ2πRln⁡x0+rx0I_{\text{loop}}=\dfrac{\mu_0 l\lambda}{2\pi R}\ln\dfrac{x_0+r}{x_0}.

Concept

The magnetic field of an infinitely long straight wire at perpendicular distance xx is

B(x)=μ0I2πx,B(x)=\frac{\mu_0 I}{2\pi x},

directed perpendicular to the loop's plane. Because BB varies across the loop, the flux must be obtained by integration.

Flux through the loop

Take a strip of width dxdx at distance xx, parallel to the wire and of length ll; its area is l dxl\,dx. Integrating from the near side x0x_0 to the far side x0+rx_0+r:

ϕ=∫x0x0+rμ0I2πx l dx=μ0Il2πln⁡ ⁣(x0+rx0).\phi=\int_{x_0}^{x_0+r}\frac{\mu_0 I}{2\pi x}\,l\,dx=\frac{\mu_0 Il}{2\pi}\ln\!\left(\frac{x_0+r}{x_0}\right).

Induced emf and current …

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