A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity

Solution:

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.