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A square loop of side $12 \mathrm{~cm}$ with its sides parallel to $X$ and $Y$ axes is moved with a velocity of $8 \mathrm{~cm} \mathrm{~s}^{-1}$ in the positive $x$-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of $10^{-3} \mathrm{~T} \mathrm{~cm}^{-1}$ along the negative $x$-direction (that is it increases by $10^{-3} \mathrm{~T} \mathrm{~cm}^{-1}$ as one moves in the negative $x$-direction). and it is decreasing in time at the rate of $10^{-3} \mathrm{Ts}^{-1}$. Determine the direction and magnitude of the induced current in the loop if its resistance $4.50 \mathrm{~m} \Omega$
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Verified Answer
Each side of square loop is $12 \mathrm{~cm}$ and magnetic field is decreasing along $x$ direction.
$$
\frac{d B}{d x}=-10^{-3} \mathrm{~T}^{\mathrm{cm}-1}=0.1 \mathrm{Tm}^{-1}
$$
also the magnetic field is decreasing with time at constant rate $\frac{d B}{d x}=-10^{-3} \mathrm{Ts}^{-1}$
Induced emf and rate of change of magnetic flux due to only time variation
$$
\begin{aligned}
&e_t=-\frac{d \phi}{d t}=-\frac{d B A}{d t}=-A \frac{d B}{d t} \\
&e_r=-0.12 \times 0.12\left[-10^{-3}\right]=144 \times 10^{-7} \mathrm{~V}
\end{aligned}
$$
Induced emf and rate of change of magnetic flux due to changes in position
$$
\begin{aligned}
e_x &=-\frac{d B A}{d t}=-A \frac{d B}{d x} \times \frac{d x}{d t} \\
e_x &=-A v \frac{d B}{d x}=-0.12 \times 0.12 \times 0.08 \times(0.1) \\
&=1152 \times 10^{-7} \mathrm{~V}
\end{aligned}
$$
Both the induced emf have same sign and thus adds to provide net Induced emf in the loop $e_{\text {net }}=e_t+e_x=1296 \times 10^{-7} \mathrm{~V}$
$$
\begin{aligned}
&\text { Induced current } \quad I=\frac{e_{\text {net }}}{R}=\frac{1296 \times 10^{-7}}{4.5 \times 10^{-3}} \\
&=2.88 \times 10^{-2} \mathrm{~A}
\end{aligned}
$$
$$
\frac{d B}{d x}=-10^{-3} \mathrm{~T}^{\mathrm{cm}-1}=0.1 \mathrm{Tm}^{-1}
$$
also the magnetic field is decreasing with time at constant rate $\frac{d B}{d x}=-10^{-3} \mathrm{Ts}^{-1}$
Induced emf and rate of change of magnetic flux due to only time variation
$$
\begin{aligned}
&e_t=-\frac{d \phi}{d t}=-\frac{d B A}{d t}=-A \frac{d B}{d t} \\
&e_r=-0.12 \times 0.12\left[-10^{-3}\right]=144 \times 10^{-7} \mathrm{~V}
\end{aligned}
$$
Induced emf and rate of change of magnetic flux due to changes in position
$$
\begin{aligned}
e_x &=-\frac{d B A}{d t}=-A \frac{d B}{d x} \times \frac{d x}{d t} \\
e_x &=-A v \frac{d B}{d x}=-0.12 \times 0.12 \times 0.08 \times(0.1) \\
&=1152 \times 10^{-7} \mathrm{~V}
\end{aligned}
$$
Both the induced emf have same sign and thus adds to provide net Induced emf in the loop $e_{\text {net }}=e_t+e_x=1296 \times 10^{-7} \mathrm{~V}$
$$
\begin{aligned}
&\text { Induced current } \quad I=\frac{e_{\text {net }}}{R}=\frac{1296 \times 10^{-7}}{4.5 \times 10^{-3}} \\
&=2.88 \times 10^{-2} \mathrm{~A}
\end{aligned}
$$
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