Question

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 move 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.

Answer 1

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,

`("dB")/("dx")` = 10⁻³ T cm⁻¹ = 10⁻¹ T m⁻¹

And, rate of decrease of the magnetic field,

`("dB")/("dt")` = 10⁻³ T s⁻¹

Resistance of the loop, R = 4.5 mΩ = 4.5 × 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:

`("d"phi)/("dt") = "A" xx ("dB")/("dx") xx "v"`

= 144 × 10⁻⁴ m² × 10⁻¹ × 0.08

= 11.52 × 10⁻⁵ T m² s⁻¹

Rate of change of the flux due to explicit time variation in field B is given as:

`("d"phi"'")/("dt") = "A" xx ("dB")/("dt")`

= 144 × 10⁻⁴ × 10⁻³

= 1.44 × 10⁻⁵ T m² 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 × 10⁻⁵ + 11.52 × 10⁻⁵

= 12.96 × 10⁻⁵ V

∴ Induced current, i = `"e"/"R"`

= `(12.96 xx 10^-5)/(4.5 xx 10^-3)`

i = 2.88 × 10⁻² A

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along the positive z-direction.