Question

A rectangular wire loop of sides $8cm$ and $2cm$ with a small cut is moving out of a region of the uniform magnetic field of magnitude $0.3T$ directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is $1cm{s}^{-1}$ in a direction normal to the (a) longer side and (b) shorter side of the loop? For how long does the induced voltage last in each case?

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Solution

**Step 1: Given**

Length of the loop, ${l}_{r}=8cm=0.08m$

Width of the loop, ${b}_{r}=2cm=0.02m$

The magnetic field of the loop, $B=0.3T$

The velocity of the loop, $v=1cm{s}^{-1}=0.01m{s}^{-1}$

**Step 2: Formula used**

Emf is in the loop, $e=Blv$

where, $B$ is the strength of the magnetic field, $l$ length of the object and $v$ is the velocity.

Distance-time-speed relationship, $\text{time}=\frac{\text{distance}}{\text{speed}}$

**Step 3: Solution**

When loop is travelling in the normal direction to its length,

$l={l}_{r}$ because it is moving along the normal to the length of the rectangle.

EMF developed in the loop,

$\begin{array}{rcl}e& =& B{l}_{r}v\\ & =& 0.3\times 0.08\times 0.01\\ & =& 2.4\times {10}^{-4}V\end{array}$

Time taken to travel along the length,

$\begin{array}{rcl}t& =& \frac{{l}_{r}}{v}\\ & =& \frac{0.08}{0.01}\\ & =& 8s\end{array}$

When loops is travelling along the normal to its breadth,

$l={b}_{r}$ because it is moving along the normal to the breadth of the rectangle.

EMF developing in the loop,

$\begin{array}{rcl}e& =& B{b}_{r}v\\ & =& 0.3\times 0.02\times 0.01\\ & =& 0.6\times {10}^{-4}V\end{array}$

Time taken to travel along the breadth,

$\begin{array}{rcl}t& =& \frac{{b}_{r}}{v}\\ & =& \frac{0.02}{0.01}\\ & =& 2s\end{array}$

**Hence, the EMFs are **$2.4\times {10}^{-4}V$**and **$0.6\times {10}^{-4}V$** and time **$8s$** and **$2s$** respectively.**

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