Question

A rod of length b moves with a constant velocity v in the magnetic field of a straight long conductor that carries a current I as shown in the figure. The emf induced in the rod is

• A. $\frac{\mu_{o}\mspace{6mu} i\mspace{6mu} v}{2\pi}\mspace{6mu}\tan^{- 1}\mspace{6mu}\frac{a}{b}$
• B. $\frac{\mu_{0}iv}{2\pi}\mathcal{\mspace{6mu} l}n\mspace{6mu}(1 + \frac{b}{a})$
• C. $\frac{\mu_{o}\mspace{6mu} i\mspace{6mu} v\mspace{6mu}\sqrt{ab}}{4\pi ⥂ (a + b)}\mspace{6mu}$
• D. $\frac{\mu_{o}\mspace{6mu} i\mspace{6mu} v\mspace{6mu}(a + b)}{4\pi\mspace{6mu} ab}\mspace{6mu}$

Answer 1

Answer: (B)

$\frac{\mu_{0}iv}{2\pi}\mathcal{\mspace{6mu} l}n\mspace{6mu}(1 + \frac{b}{a})$

Answer 2

The induced emf between two ends of a segment dx = dE = Bvdx

∴ where B = magnetic field due to straight current carrying

wire

at the segment dx = $\frac{\mu_{0}i}{2\pi x}$ ⇒ dE = $\frac{\mu_{0}ivdx}{2\pi\mspace{6mu} x}$

∴ The induced emf between the ends of the rod

= E = $\int_{}^{}{dE}\mspace{6mu} = \mspace{6mu}\frac{\mu_{0}iv}{2\pi}\mspace{6mu}\int_{\mspace{6mu} a}^{\mspace{6mu} a + b}{\mspace{6mu}\frac{dx}{x}}$

⇒ E = $\frac{\mu_{0}iv}{2\pi\mspace{6mu}}\mspace{6mu}\ln\mspace{6mu}\left( \frac{b}{a} + 1 \right)$ Hence (2) is correct.