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: GivenLength 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 usedEmf 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: SolutionWhen 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,$\begin{array}{rcl}e& =& B{l}_{r}v\\ & =& 0.3×0.08×0.01\\ & =& 2.4×{10}^{-4}V\end{array}$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,$\begin{array}{rcl}e& =& B{b}_{r}v\\ & =& 0.3×0.02×0.01\\ & =& 0.6×{10}^{-4}V\end{array}$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 $2.4×{10}^{-4}V$and $0.6×{10}^{-4}V$ and time $8s$ and $2s$ respectively.

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