Electromagnetic Induction Question 1
Question 1 - 2024 (01 Feb Shift 1)
A rectangular loop of sides $12 cm$ and $5 cm$, with its sides parallel to the $x$-axis and $y$-axis respectively moves with a velocity of $5 cm / s$ in the positive $x$ axis direction, in a space containing a variable magnetic field in the positive $z$ direction. The field has a gradient of $10^{-3} T / cm$ along the negative $x$ direction and it is decreasing with time at the rate of $10^{-3} T / s$. If the resistance of the loop is $6 m \Omega$, the power dissipated by the loop as heat is $\times 10^{-9} W$.
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Answer: (216)
Solution:
$B _0$ is the magnetic field at origin
$\frac{d B}{d x}=-\frac{10^{-3}}{10^{-2}}$
$\int _{B _0}^{B} d B=-\int _0^{x} 10^{-1} d x$
$B-B _0=-10^{-1} x$
$B=\left(B _0-\frac{x}{10}\right)$
Motional emf in $AB=0$
Motional emf in $CD=0$
Motional emf in $AD=\varepsilon _1=B _0 \ell v$
Magnetic field on rod BC B
$=\left(B _0-\frac{\left(-12 \times 10^{-2}\right)}{10}\right)$
Motional emf in $BC=\varepsilon _2=\left(B _0+\frac{12 \times 10^{-2}}{10}\right) \ell \times v$
$\varepsilon _{\text {eq }}=\varepsilon _2-\varepsilon _1=300 \times 10^{-7} V$
For time variation
$$ \begin{aligned} & \left(\varepsilon _{eq}\right)^{\prime}=A \frac{dB}{dt}=60 \times 10^{-7} V \\ & \left(\varepsilon _{eq}\right) _{\text {net }}=\varepsilon _{eq}+\left(\varepsilon _{eq}\right)^{\prime}=360 \times 10^{-7} V \\ & \text { Power }=\frac{\left(\varepsilon _{eq}\right) _{\text {net }}^{2}}{R}=216 \times 10^{-9} W \end{aligned} $$
