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 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 mW.

Answer 1

A square loop of side length is shown in figure. here magnetic field is decreasing along x - direction.

so, $\frac{dB}{dx}=-10^{-3}Tcm^{-1}=-0.1Tm^{-1}$ [ as we know, 1 m = 100cm ]

also the magnetic field is decreasing with time at constant rate,$\frac{dB}{dt}=-10^{-3}Ts^{-1}$

so, induced emf and rate of change of magnetic flux due to only time variation is ,$E_t=-\frac{d\phi}{dt}$

= $-\frac{dBA}{dt}=-A\frac{dB}{dt}$

here A =area of square loop = (12cm)² = 0.0144 m²

so, $E_t=-(0.0144)\times(-10^{-3})$

= 144 × 10^-7 V

induced emf and the rate of change of magnetic flux due to change in position.

$E_x=-\frac{dBA}{dt}=-A\frac{dB}{dx}\times\frac{dx}{dt}$

here v is velocity of square loop .i.e., 8cm/s

= $-Av\frac{dB}{dx}$

= -(0.0144) × (8 × 10^-2) × (- 0.1 )

= 1152 × 10^-7V

both the induced emf have same sign and thus adds to provide net induced emf in the loop.

so, E = $E_t+E_x$

= (144 + 1152) × 10^-7 V

= 1296 × 10^-7 V = 1.296 × 10^-4 V

and induced current , I = E/R

here R = 4.5 mΩ = 4.5 × 10^-3 Ω

= 1.296 × 10^-4/4.5 × 10^-3

= 2.88 × 10^-2 A

Answer 2

Answer:

A square loop of side length is shown in figure. here magnetic field is decreasing along x - direction.

so, \frac{dB}{dx}=-10^{-3}Tcm^{-1}=-0.1Tm^{-1}dxdB=−10−3Tcm−1=−0.1Tm−1 [ as we know, 1 m = 100cm ]

also the magnetic field is decreasing with time at constant rate,\frac{dB}{dt}=-10^{-3}Ts^{-1}dtdB=−10−3Ts−1

so, induced emf and rate of change of magnetic flux due to only time variation is ,E_t=-\frac{d\phi}{dt}Et=−dtdϕ

= -\frac{dBA}{dt}=-A\frac{dB}{dt}−dtdBA=−AdtdB

here A =area of square loop = (12cm)² = 0.0144 m²

so, E_t=-(0.0144)\times(-10^{-3})Et=−(0.0144)×(−10−3)

= 144 × 10^-7 V

induced emf and the rate of change of magnetic flux due to change in position.

E_x=-\frac{dBA}{dt}=-A\frac{dB}{dx}\times\frac{dx}{dt}Ex=−dtdBA=−AdxdB×dtdx

here v is velocity of square loop .i.e., 8cm/s

= -Av\frac{dB}{dx}−AvdxdB

= -(0.0144) × (8 × 10^-2) × (- 0.1 )

= 1152 × 10^-7V

both the induced emf have same sign and thus adds to provide net induced emf in the loop.

so, E = E_t+E_xEt+Ex

= (144 + 1152) × 10^-7 V

= 1296 × 10^-7 V = 1.296 × 10^-4 V

and induced current , I = E/R

here R = 4.5 mΩ = 4.5 × 10^-3 Ω

= 1.296 × 10^-4/4.5 × 10^-3

= 2.88 × 10^-2 A