A copper rod of length $$0.19\,m$$  is moving g parallel to a long wire with a uniform velocity of $$10\,m/s.$$  The long wire carries 5 ampere current and is perpendicular to the rod. The ends of the rod are at distances $$0.01\,m$$  and $$0.2\,m$$  from the wire. The emf induced in the rod will be -

Solution :
EMF induced in an element of length $$dx$$ at a distance $$x$$ from wire $$= Bvdx$$
∴ Total EMF induced in the rod
$$\eqalign{ & E = \int\limits_{0.01}^{0.2} {Bv} \,dx = \int\limits_{0.01}^{0.2} {\frac{{{\mu _0}iv}}{{2\pi x}}} dx = \frac{{{\mu _0}iv}}{{2\pi }}\int\limits_{0.01}^{0.2} {\frac{1}{x}} dx \cr & E = \frac{{{\mu _0}iv}}{{2\pi }}\left[ {{{\log }_e}x} \right]_{0.01}^{0.2} = \frac{{{\mu _0}iv}}{{2\pi }}\left[ {{{\log }_{10}}\left( {0.2} \right) - {{\log }_{10}}\left( {0.01} \right)} \right] \times 2.303 \cr & E = \frac{{4\pi \times {{10}^{ - 7}} \times 5 \times 10}}{{2\pi }}\left[ {1.301} \right] \times 2.303 \cr & = 2.99 \times {10^{ - 5}}V \cr & \approx 30\,\mu V \cr} $$