(ii) Now, consider that we passed an electric current through the straight wire of 50 A, and the loop is then moved to the right with constant velocity, v = 10 m/s.Find the emf induced in the loop at an instant where x = 0.2 m. Take a = 0.01 m and assume that the loop has a large resistance.

Answer –

(i) In the loop, place a small element dy at a distance y from the long straight wire (as shown in the given figure).

Magnetic flux associated with element dy is given by dϕ = ady

B represents the magnetic field at distance, given by  

y = 2πyμ₀​I​

where, I is the current in the wire

μ_0​ = Permeability of free space = 4π×10^-7

Therefore,

\[\therefore d\phi =\frac{{{\mu }_{0}}Ia}{2\pi }\frac{dy}{y}\]

\[\phi =\frac{{{\mu }_{0}}Ia}{2\pi }\int\limits_{x}^{x+a}{\frac{dy}{y}}\]

\[\phi =\frac{{{\mu }_{0}}Ia}{2\pi }\left[ {{\log }_{e}}y \right]_{x}^{a+x}\]

\[\phi =\frac{{{\mu }_{0}}Ia}{2\pi }{{\log }_{e}}\frac{a+x}{x}\]

For mutual inductance M, the flux is given as –

\[\phi =MI\]

\[\therefore MI=\frac{{{\mu }_{0}}Ia}{2\pi }{{\log }_{e}}\left( \frac{a}{x}+1 \right)\]

\[MI=\frac{{{\mu }_{0}}a}{2\pi }{{\log }_{e}}\left( \frac{a}{x}+1 \right)\]

(ii) EMF induced in the loop is given by e = B’av, therefore –

\[e=\frac{{{\mu }_{0}}I}{2\pi x}av\]

Given, I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s

\[e=\frac{4\pi \times {{10}^{-7}}\times 50\times 0.1\times 10}{2\pi \times 0.2}\]

\[e=5\times {{10}^{-5}}V\]