Answer to Question #109423 in Electricity and Magnetism for happy45

Question

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R=4

Answer 1

Question

A rectangular conducting loop with N=100 turns is placed below a long straight conducting wire with current of 10 A as shown in the figure. The loop is moving at a speed of v = 6 cm/s to the -j direction. The distance x is measured between the wire and the center axis of the loop.

(a) State the direction of magnetic field through the loop as produced by the wire.

(b) Show that the total magnetic flux through the loop is

$$\Phi_B=\frac{\mu_0iL}{2\pi}\text{ ln}\frac{x+\frac{W}{2}}{x-\frac{W}{2}}.$$

(c) Given x = 10 cm, L = 20 cm, W = 6 cm, and the loop resistance is (R+8) Ω, find the magnitude and direction of the induced current.

Solution

(a) According right-hand rule, the direction of magnetic field through the loop is in -k direction.

(b) The total magnetic flux is Flux density of magnetic field times Area of the loop. The flux density varies along the W side of the loop because it is further from the center of the wire. Consider a small width segment dx along W. The magnetic field B, therefore, will be

$$B(x)=\frac{\mu_0i}{2\pi x}.$$ $$d\Phi=B(x)Ldx=\frac{\mu_0i}{2\pi x}Ldx.$$

Integrate it from x-W/2 to x+W/2:

$$\Phi=\int_{x-\frac{W}{2}}^{x+\frac{W}{2}}d\Phi=\frac{\mu_0iL}{2\pi}\int_{x-\frac{W}{2}}^{x+\frac{W}{2}}\frac{dx}{x}=\\ \space\\ =\frac{\mu_0iL}{2\pi}\text{ ln}\frac{x+\frac{W}{2}}{x-\frac{W}{2}}.$$

(c) Find the induced EMF:

$$E=-N\frac{d\Phi}{dt}=-N\frac{\mu_0iL}{2\pi}\bigg(\frac{1}{x+\frac{W}{2}}-\frac{1}{x-\frac{W}{2}}\bigg)\frac{dx}{dt}=\\ \space\\ =-N\frac{\mu_0iL}{2\pi}\bigg(\frac{1}{x+\frac{W}{2}}-\frac{1}{x-\frac{W}{2}}\bigg)v.$$

According to Ohm's law, the current is

$$I=\frac{E}{R}=N\frac{\mu_0ivL}{2\pi R}\bigg(\frac{1}{x+\frac{W}{2}}-\frac{1}{x-\frac{W}{2}}\bigg)=\\ \space\\ =1.32\cdot10^{-6}\text{ A}.$$