Class 12 Physics
CBSE and UP Board
Electromagnetic Induction
Chapter-6 Exercise 6.12
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 6 Question 12
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 10⁻³T cm⁻¹ along the negative x-direction (that is it increases by 10⁻³T cm⁻¹ as one moves in the negative x-direction), and it is decreasing in time at the rate of 10⁻³T s⁻¹. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ..
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Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, A = 0.12 x 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,
dB/dx = 10⁻³ Tcm⁻¹= 10⁻¹ T m⁻¹
And, rate of decrease of the magnetic field,
dB/dt = 10⁻³ Ts⁻¹
Resistance of the loop, R = 4.5 mΩ = 4.5 x 10-3 Ω
Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:
dφ/dt =A x dB/dx x v
= 144 x 10⁻⁴m² x 10⁻¹ x 0.08
=11.52 x 10⁻⁵ Tm² s⁻¹
Rate of change of the flux due to explicit time variation in field B is given as:
dφ´/dt =A x dB/dt
= 144 x 10⁻⁴ x 10⁻³
= 1.44 x 10⁻⁵ Tm² 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 x 10⁻⁵+ 11.52 x 10⁻⁵
= 12.96 x 10⁻⁵ V
Therefore, Induced current, i = e/R
= (12.96 x 10⁻⁵)/(4.5 x 10-3) = 2.88 x 10⁻² 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.