Question

A long solenoid of radius $R$ carries a time $\left(t\right)$ dependent current $I\left(t\right)=I_{0}t(1-t)$. A ring of radius $2R$ is placed coaxially near its middle. During the time instant $0\le t\le 1$, the induced current $\left(I_R\right)$ and the induced$EMF\left(V_R\right)$ in the ring changes as:

• A. Direction of $IR$ remains unchanged and $V_R$ is maximum at$t=0.5$
• B. Direction of$IR$ remains unchanged and $V_R$ is zero at$t=0.25$
• C. At $t=0.5$ direction of $IR$ reverses and $V_R$ is zero
• D. At $t=0.25$ direction of $IR$ reverses and $V_R$ is maximum

Answer 1

The correct option is C

At $t=0.5$ direction of $IR$ reverses and $V_R$ is zero

Step 1: Given data

Current $I\left(t\right)=I_{0}t(1-t)$

The radius of the ring $=2R$

The radius of the solenoid $=R$

Step 2: Find out the flux, $\mathit{\varphi }$

Field due to solenoid near the middle, $B={\mu }_{o}NI$ (Where, $N=$number of turns per unit length)

Flux, $\varphi =B.A$ [Where $A={\mathrm{\pi R}}^2$(cross-sectional area)]

$$\begin{gathered} ={\mu }_{o}NI.\pi R^2 \\ ={\mu }_{o}N{I}_{o}t(1-t).\pi R^2 \end{gathered}$$

Step 3: Find out the relation between the induced current $\left(I_R\right)$ and the induced$EMF\left(V_R\right)$ in the ring and draw the diagram

Induced emf, $V_R=-d\varphi /dt$ [By Lenz’s law]

$$\begin{gathered} =-\frac{d{\mu }_{o}N{I}_{o}t(1-t).\pi R^2}{dt} \\ =-\pi {\mu }_o{I}_{o}NR2(1-2t) \end{gathered}$$

As, $V=IR$ And we can see $V_R$ is zero for

$$\begin{gathered} \Rightarrow (1-2t)=0 \\ \Rightarrow t=(1/2)s \end{gathered}$$

$V_R$ is negative for $t<1/2$ and increasing linearly

$V_R$ is positive for $t>1/2$ and increasing linearly

The same will be true for current, $I[As,V\propto R]$

Hence, option (C) is correct.