$A$ square loop of side $12 \; cm$ with its sides parallel to $X$ and $Y$ axes is moved with a velocity of $8 \; cm \, s^{-1}$ 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 $10^{-3} \; T \, cm^{-1}$ along the negative $x$-direction (that is, it increases by $10^{-3} \; T \, cm^{-1}$ as one moves in the negative $x$-direction), and it is decreasing in time at the rate of $10^{-3} \; T \, s^{-1}$. Determine the direction and magnitude of the induced current in the loop if its resistance is $4.50 \; m\Omega$.
$(2.88 \times 10^{-2} \; A)$ Side of the square loop, $s = 12 \; cm = 0.12 \; m$.
Area of the square loop, $A = 0.12 \times 0.12 = 0.0144 \; m^2$.
Velocity of the loop, $v = 8 \; cm/s = 0.08 \; m/s$.
Gradient of the magnetic field along the negative $x$-direction, $\frac{dB}{dx} = 10^{-3} \; T \, cm^{-1} = 10^{-1} \; T \, m^{-1}$.
Rate of decrease of the magnetic field, $\frac{dB}{dt} = 10^{-3} \; T \, s^{-1}$.
Resistance of the loop, $R = 4.5 \; m\Omega = 4.5 \times 10^{-3} \; \Omega$.
The rate of change of magnetic flux due to the motion of the loop in a non-uniform magnetic field is $\frac{d\phi_1}{dt} = A \times \frac{dB}{dx} \times v = 0.0144 \times 10^{-1} \times 0.08 = 1.152 \times 10^{-4} \; Wb/s$.
The rate of change of flux due to the explicit time variation of the field is $\frac{d\phi_2}{dt} = A \times \frac{dB}{dt} = 0.0144 \times 10^{-3} = 0.144 \times 10^{-4} \; Wb/s$.
The total induced emf is $e = \frac{d\phi_1}{dt} + \frac{d\phi_2}{dt} = 1.152 \times 10^{-4} + 0.144 \times 10^{-4} = 1.296 \times 10^{-4} \; V$.
The induced current is $i = \frac{e}{R} = \frac{1.296 \times 10^{-4}}{4.5 \times 10^{-3}} = 2.88 \times 10^{-2} \; A$.
By Lenz's law, the direction of the current is such that it opposes the change in flux. Since the flux is increasing due to motion and decreasing due to time variation, the net effect results in a clockwise current when viewed from the positive $z$-axis.
Area of the square loop, $A = 0.12 \times 0.12 = 0.0144 \; m^2$.
Velocity of the loop, $v = 8 \; cm/s = 0.08 \; m/s$.
Gradient of the magnetic field along the negative $x$-direction, $\frac{dB}{dx} = 10^{-3} \; T \, cm^{-1} = 10^{-1} \; T \, m^{-1}$.
Rate of decrease of the magnetic field, $\frac{dB}{dt} = 10^{-3} \; T \, s^{-1}$.
Resistance of the loop, $R = 4.5 \; m\Omega = 4.5 \times 10^{-3} \; \Omega$.
The rate of change of magnetic flux due to the motion of the loop in a non-uniform magnetic field is $\frac{d\phi_1}{dt} = A \times \frac{dB}{dx} \times v = 0.0144 \times 10^{-1} \times 0.08 = 1.152 \times 10^{-4} \; Wb/s$.
The rate of change of flux due to the explicit time variation of the field is $\frac{d\phi_2}{dt} = A \times \frac{dB}{dt} = 0.0144 \times 10^{-3} = 0.144 \times 10^{-4} \; Wb/s$.
The total induced emf is $e = \frac{d\phi_1}{dt} + \frac{d\phi_2}{dt} = 1.152 \times 10^{-4} + 0.144 \times 10^{-4} = 1.296 \times 10^{-4} \; V$.
The induced current is $i = \frac{e}{R} = \frac{1.296 \times 10^{-4}}{4.5 \times 10^{-3}} = 2.88 \times 10^{-2} \; A$.
By Lenz's law, the direction of the current is such that it opposes the change in flux. Since the flux is increasing due to motion and decreasing due to time variation, the net effect results in a clockwise current when viewed from the positive $z$-axis.
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