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A small conducting loop of radius a and resistance r is pulled with velocity V perpendicular to a long straight conductor carrying a current i as shown in fig the current I in terms of V is

Option: 1
$\frac{\mu _0ia^2}{2x^2}V$

Option: 2
$\frac{\mu _0ia^2V^2}{2x^2r}$

Option: 3
$\frac{\mu _0ia^2}{2x^2r^2}V$

Option: 4
$\frac{\mu _0ia^2V}{2x^2r}V$

Answers (1)

best_answer

As we have learned

Induced Current -

$I= \frac{\varepsilon }{R}=\frac{-N}{R}\frac{d\phi }{dt}$

- wherein

$R\rightarrow$ Resistance

$\frac{d\phi }{dt}\rightarrow$ Rate of change of flux

MAgnetic field at position of loop

$B = \frac{\mu _0i}{2xr}$

Since loop is small , magnetic field can be assumed to be constant .

So, flux across loop $\phi = \int \vec B \cdot d \vec A$

$= \vec B \int d \vec A$

$= BA$

= $\frac{\mu _0i \pi a^2}{2 \pi x}$

$\frac{\mu _0i a^2}{2 x}$

Now induced enf ,

$\varepsilon = \frac{-d\phi }{dt }$

$= \frac{-\mu _0ia^2}{2}\left ( \frac{d(1/x)}{dt} \right )$

$= \frac{\mu _0ia^2}{2x^2}\left ( \frac{d(x)}{dt} \right )$

$=\frac{\mu _0ia^2}{2x^2}V$

Now induced current

$I = \frac{\varepsilon }{r}$

$I = \frac{\mu _0ia^2}{2x^2r}V$

Posted by

avinash.dongre

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