Question

A horizontal telegraph wire $0.5km$ long running east and west in a part of a circuit whose resistance is $2.5\Omega$. The wire falls to $g=10.0m/s^2$ and $B=2\times {10}^{-5}weber/m^2$, then the current induced in the circuit is

• A. $0.7amp$
• B. $0.04amp$
• C. $0.02amp$
• D. $0.01amp$

Answer 1

The correct option is C $0.02amp$

$i=\frac{\epsilon }{R}=\frac{1}{R}\frac{d\varphi }{dt}$

Here $df=B\times A=(2\times {10}^{-5})\times 0.5\times {10}^{+3}\times 5)$

$dt=$ time taken by the wire to fall at ground

$=(2h/g{)}^{1/2}=(10/10{)}^{1/2}=1sec$

$\therefore i=\frac{1}{2.5}\left[\frac{(2\times {10}^{-5})\times (0.5\times {10}^3\times 5)}{1}\right]=0.02amp$.