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A rectangular wire loop of sides $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \mathrm{~T}$ directed normal to the loop. (a) What is the emf developed across the cut if the velocity of the loop is $1 \mathrm{~cm} \mathrm{~s}^{-1}$ in a direction normal to the (1) longer side, (2) shorter side of the loop? (b) For how long does the induced voltage last in each case?
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Verified Answer
Given: Magnetic field $\mathrm{B}=0.3 \mathrm{~T}$
velocity of loop $=1 \mathrm{~cm} / \mathrm{s}=10^{-2} \mathrm{~m} / \mathrm{s}$
area of loop $=8 \mathrm{~cm} \times 2 \mathrm{~cm}$
To find: (a) Voltage in the cut for (1) long side moving out
(2) shorter side
(b) time of emf. for each case.
Formula: (a) $\mathrm{e}=\mathrm{B} \ell \mathrm{v}$ (b) $\mathrm{t}=\frac{\text { length of wire }}{\mathrm{v}}$
(i) along longer side
length $=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$
$\therefore \mathrm{e}=\mathrm{B} \ell \mathrm{v}=0.3 \times 8 \times 10^{-2} \times 10^{-2}=0.24 \times 10^{-3} \mathrm{~V}=0.24 \mathrm{mV}$
Time of emf $=\frac{\text { length of shorter arm }}{\mathrm{v}}$
( $\because$ emf develops as long as loop doesnot get out of the field ie, distance travelled by shorter arm)
$$
\mathrm{t}=\frac{2 \times 10^{-2}}{10^{-2}}=2 \times 10^{-2+2}=2 \mathrm{sec} .
$$
(ii) along shorter side
$$
\begin{array}{r}
\mathrm{e}=\mathrm{B} \ell \mathrm{v}=0.3 \times 2 \times 10^{-2} \times 10^{-2} \\
=0.6 \times 10^{-4}=0.06 \mathrm{mV}
\end{array}
$$
time of $\operatorname{emf}=\frac{\text { length of longer arm }}{\mathrm{v}}$
$$
=\frac{8 \times 10^{-2}}{10^{-2}}=8 \mathrm{sec}
$$
velocity of loop $=1 \mathrm{~cm} / \mathrm{s}=10^{-2} \mathrm{~m} / \mathrm{s}$
area of loop $=8 \mathrm{~cm} \times 2 \mathrm{~cm}$
To find: (a) Voltage in the cut for (1) long side moving out
(2) shorter side
(b) time of emf. for each case.
Formula: (a) $\mathrm{e}=\mathrm{B} \ell \mathrm{v}$ (b) $\mathrm{t}=\frac{\text { length of wire }}{\mathrm{v}}$
(i) along longer side
length $=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$
$\therefore \mathrm{e}=\mathrm{B} \ell \mathrm{v}=0.3 \times 8 \times 10^{-2} \times 10^{-2}=0.24 \times 10^{-3} \mathrm{~V}=0.24 \mathrm{mV}$
Time of emf $=\frac{\text { length of shorter arm }}{\mathrm{v}}$
( $\because$ emf develops as long as loop doesnot get out of the field ie, distance travelled by shorter arm)
$$
\mathrm{t}=\frac{2 \times 10^{-2}}{10^{-2}}=2 \times 10^{-2+2}=2 \mathrm{sec} .
$$
(ii) along shorter side
$$
\begin{array}{r}
\mathrm{e}=\mathrm{B} \ell \mathrm{v}=0.3 \times 2 \times 10^{-2} \times 10^{-2} \\
=0.6 \times 10^{-4}=0.06 \mathrm{mV}
\end{array}
$$
time of $\operatorname{emf}=\frac{\text { length of longer arm }}{\mathrm{v}}$
$$
=\frac{8 \times 10^{-2}}{10^{-2}}=8 \mathrm{sec}
$$
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