Question

(A) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21. (B) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Answer 1

The induced emf in the coil is e = 5×10^−5 V

Explanation:

Take a small element dy in the loop at a distance y from the long straight wire.

Magnetic flux associated with element, dy is dϕ =BdA

Where, dA = a×dy is Area of element dy

B= μoI /2πy is the magnetic field at y.

I is current in the wire.

dϕ = μo Ia× dy/2πy

ϕ =  (μo Ia/2π)∫  a+x - x  dy/y

ϕ = (μo Ia / 2π) ln (  a+x/x  )

Mutual Inductance "M" = ϕ/I = (μo a/2π)ln(  a+x /x)

Emf induced in the loop, e= Bav = (μo I/2πx)av

From the given values, e = 5×10^−5 V

Thus the induced emf in the coil is e = 5×10^−5 V

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Answer 2

(A) Expression for the mutual inductance is:

$\bold{M=\frac{\mu_0a}{2\pi}\left [ log_e \left ( \frac{a}{x}+1 \right ) \right ]}$

(B) The induced emf in the loop at the instant when x = 0.2 m is $\bold{5\times 10^{-5} \ V}$

Solution:

(A) Expression for the mutual inductance:

Consider the diagram given in the question.

Magnetic field due to current at a distance x from wire is:

$\bold{B=\frac{\mu_0 I}{2\pi x}}$

Let us consider small area 'dA' at a distance 'dx' from the wire. The magnetic flux at that portion is given by the formula:

$\bold{d\phi = B.dA}$

$\bold{d\phi=\frac{\mu_0 I}{2\pi x}(a.dx)}$

Magnetic flux linked with the square loop is given as:

$\bold{\phi=\frac{\mu_0Ia}{2\pi}\int_{x}^{x+a}\frac{dx}{x}}$

$\bold{\phi=\frac{\mu_0Ia}{2\pi}\left [ log_eX \right ]_{x}^{x+a}}$

$\bold{\phi=\frac{\mu_0Ia}{2\pi}\left [ log_e \left ( \frac{x+a}{x} \right ) \right ]}$

$\bold{\phi=\frac{\mu_0Ia}{2\pi}\left [ log_e \left ( \frac{a}{x}+1 \right ) \right ]}$

Now, the flux and mutual inductance is given by the formula:

$\bold{\phi=MI}$

On equating flux and mutual inductance, we get,

$\bold{MI=\frac{\mu_0Ia}{2\pi}\left [ log_e \left ( \frac{a}{x}+1 \right ) \right ]}$

$\bold{M=\frac{\mu_0a}{2\pi}\left [ log_e \left ( \frac{a}{x}+1 \right ) \right ]}$

An expression for the mutual inductance between a long straight wire and a square loop is obtained.

(B) The induced emf in the loop at the instant when x = 0.2 m:

Induced emf formula is:

$\bold{e=BLv}$

Where,

L = Current = 0.1 m

v = Velocity = 10 m/s

B = Magnetic field Induction $\bold{=\frac{\mu_0I}{2\pi x}}$

Where,

$\bold{\mu_0}$ = Vacuum permeability $\bold{=4\pi\times10^{-7} \ H/m}$

I = Current = 50 A

x = Distance = 0.2 m

Now, on substituting the values, we get,

$\bold{e=(\frac{\mu_0I}{2\pi x})Iv}$

$\bold{e=\frac{4\pi\times10^{-7}\times50\times0.1\times10}{2\pi\times0.2}}$

$\bold{\therefore e=5\times10^{-5} \ Volt}$