Question

A current, $i=3.36(1+2t){10}^{-2}\text{A}$ increases at a steady rate in a long straight wire. A small circular loop of radius ${10}^{-3}\text{m}$ has its plane parallel to the wire and is placed at a distance of $1\text{m}$ from the wire. The resistance of the loop is $8.4\times {10}^{-4}\Omega$. Find the magnitude of induced current in the loop.

• A. $1\times {10}^{-11}\text{A}$
• B. $2.5\times {10}^{-11}\text{A}$
• C. $5\times {10}^{-11}\text{A}$
• D. $1.5\times {10}^{-11}\text{A}$

Answer 1

The correct option is C $5\times {10}^{-11}\text{A}$

The situation described in question is shown in figure,


Magnetic field at the centre of coil due to long straight wire is,

$B=\frac{{\mu }_{0}i}{2\pi r}$

The magnitude flux passing through the loop is given as,

$\varphi =BA\cos 0^{\circ }=\left(\frac{{\mu }_{0}i}{2\pi r}\right)\pi a^2=\frac{{\mu }_0{a}^{2}i}{2r}$

Substituting the value of $i$,

$\varphi =\frac{{\mu }_0{a}^2}{2\times 1}\left[3.36\right(1+2t)\times {10}^{-2}]$

The magnitude of induced $\text{emf}$ is given by,

$E=\frac{d\varphi }{dt}=\frac{{\mu }_0{a}^2}{2}\times [3.36\times 2\times {10}^{-2}]$

$\therefore E=3.36\times {10}^{-2}\times {\mu }_0{a}^2$

So, the induced current $i_{ind}$ is given by,

$i_{ind}=\frac{E}{R}=\frac{1}{R}\times 3.36\times {10}^{-2}\times {\mu }_0{a}^2$

Substituting the values,

$i_{ind}=\frac{(3.36\times {10}^{-2})\times (4\pi \times {10}^{-7})({10}^{-3}{)}^2}{(8.4\times {10}^{-4})}$

$=\frac{4.22\times {10}^{-14}}{8.4\times {10}^{-4}}\approx 5\times {10}^{-11}\text{A}$

Hence, option $\left(C\right)$ is the correct answer.