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 mΩ.
Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, A = 0.12 × 0.12 = 0.0144 m2
Velocity of the loop, v = 8 cm/s = 0.08 m/s
Gradient of the magnetic field along negative x-direction,
$\frac{d B}{d x}=10^{-3} \mathrm{Tcm}^{-1}=10^{-1} \mathrm{Tm}^{-1}$
And, rate of decrease of the magnetic field,
$\frac{d B}{d t}=10^{-3} \mathrm{~T} \mathrm{~s}^{-1}$
Resistance of the loop, $R=4.5 \mathrm{~m} \Omega=4.5 \times 10^{-3} \Omega$
Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:
$\frac{d \phi}{d t}=A \times \frac{d B}{d x} \times v$
$=144 \times 10^{-4} \mathrm{~m}^{2} \times 10^{-1} \times 0.08$
$=11.52 \times 10^{-5} \mathrm{Tm}^{2} \mathrm{~s}^{-1}$
Rate of change of the flux due to explicit time variation in field B is given as:
$\frac{d \phi^{\prime}}{d t}=A \times \frac{d B}{d t}$
$=144 \times 10^{-4} \times 10^{-3}$
$=1.44 \times 10^{-5} \mathrm{~T} \mathrm{~m}^{2} \mathrm{~s}^{-1}$
Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:
$e=1.44 \times 10^{-5}+11.52 \times 10^{-5}$
$=12.96 \times 10^{-5} \mathrm{~V}$
$\therefore$ Induced current, $i=\frac{e}{R}$
$=\frac{12.96 \times 10^{-5}}{4.5 \times 10^{-3}}$
$i=2.88 \times 10^{-2} \mathrm{~A}$
Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.