Question

An infinitely long straight wire carrying current $I$, one side opened rectangular loop and a conductor $C$ with a sliding connector are located in the same plane, as shown in the figure. The connector has length $l$ and resistance $R$. It slides to the right with a velocity $v$. The resistance of the conductor and the self-inductance of the loop are negligible. The induced current in the loop, as a function of separation $r$, between the connector and the straight wire is:

• A. $\frac{{\mu }_0}{2\pi }\left(\frac{Ivl}{rR}\right)$
• B. $\frac{{\mu }_0}{\pi }\left(\frac{Ivl}{rR}\right)$
• C. $\frac{2{\mu }_0}{\pi }\left(\frac{Ivl}{rR}\right)$
• D. $\frac{{\mu }_0}{4\pi }\left(\frac{Ivl}{rR}\right)$

Answer 1

The correct option is A

$\frac{{\mu }_0}{2\pi }\left(\frac{Ivl}{rR}\right)$

Step1: Given data.

Carrying current in infinite long wire$=I$

Connector length$=l$

Connector resistance$=R$

Sliding velocity$=v$

Separation distance between connector and infinite wire$=r$

Step2: Finding the induced current in the wire.

Since,

Field, due to the current carrying wire in the region, right to it, is directed into the plane of the paper and its magnitude is given by,

$B=\frac{{\mu }_{0}I}{2\pi r}$ …$\left(i\right)$

Also, B is same along the length of the rod thus motional induced $e.m.f$.

$e=vBl$ ….$\left(ii\right)$

Now, we know that,

Induced current $i_{in}=\frac{e}{R}$

$i_{in}=\frac{vBl}{R}$ $\left[Fromequation\left(i\right)\right]$

$i_{in}=\frac{{\mu }_{0}ivl}{2\pi rR}$ $\left[Fromequation\left(ii\right)\right]$

$i_{in}=\frac{{\mu }_0}{2\pi }\left(\frac{Ivl}{rR}\right)$

Hence, option A is correct.