Question

Consider a long wire carrying a time varying current $i=kt(k>0)$. A circular loop of radius $a$ and resistance $R$ is placed with its center at a distance $d$ from the wire $(a<<d)$. The induced current in the loop is

• A. $\frac{{\mu }_0{a}^{2}k}{4dR}$
• B. $\frac{{\mu }_0{d}^{2}k}{2aR}$
• C. $\frac{{\mu }_0{a}^{2}k}{2dR}$
• D. ${\mu }_0\frac{(a+d{)}^{2}k}{2aR}$

Answer 1

The correct option is C $\frac{{\mu }_0{a}^{2}k}{2dR}$

Since, current in the wire is continuously increasing, we can conclude that the magnetic field due to this wire in the region, is also increasing with time.

Since, $a<<d$, we can assume the magnetic field to be constant throughout the loop.

Magnetic field $B$ due to wire at the location of loop is given by,

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

Directed perpendicular to the plane of paper and inwards.

$\therefore$ flux through the circular loop is,

$\varphi =\frac{{\mu }_{0}i}{2\pi d}\times \pi a^2$

$\varphi =\frac{{\mu }_0{a}^{2}kt}{2d}$

$\therefore$ Induced emf in the loop is given by,

$E=\left(\frac{d\varphi }{dt}\right)$

$E=\frac{d}{dt}\left(\frac{{\mu }_0{a}^{2}kt}{2d}\right)=\frac{{\mu }_0{a}^{2}k}{2d}$

Hence, the induced current in the loop is,

$i=\frac{E}{R}=\frac{{\mu }_0{a}^{2}k}{2dR}$

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

Why this Question? Key point: Since the current is a function of time, the $\overrightarrow{B}$ through the loop increases with time, and hence the flux through the loop changes. If the current through the wire would have been constant, the induced emf and the induced current through the loop would have been zero.