6.12 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 10–3 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.

6.12 Side of the square loop, s = 12 cm = 0.12 mArea of the square loop, A = 0.12 ×0.12=0.0144m2Velocity of the loop, v = 8 cm /s = 0.08 m/sGradient of the magnetic field along negative x – direction.dBdx = 10-3 T/cm = 10-1 T/mRate of decrease of magnetic fielddBdt = 10-3 T/sResistance of the loop, R = 4.50 mΩ = 4.5 ×10-3 ΩRate of change of the magnetic flux due to motion of the loop in a non-uniform magnetic field is given as:d∅dt = A ×dBdx ×v = 0.0144×10-1×0.08 = 1.152 ×10-4 T m2s-1Rate of change of flux due to explicit time variation in field B is given as:d∅'dt = A ×dBdt = 0.0144×10-3 = 1.44 ×10-5 T m2s-1Since the rate of change of the flux is the induced emf, the total emf in the loop can be calculated as:e = 1.152 ×10-4 + 1.44 ×10-5 = 1.296 ×10-4 VThe induced current, I = eR = 1.296×10-44.5×10-3 = 28.8 ×10-3 ATherefore, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

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Electromagnetic Induction Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Electromagnetic Induction

6.12 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^–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.

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Payal Gupta | Contributor-Level 10

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6.12 Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 $\times 0.12 = 0.0144 m^{2}$

Velocity of the loop, v = 8 cm /s = 0.08 m/s

Gradient of the magnetic field along negative x – direction.

$\frac{d B}{d x}$ = $10^{- 3}$ T/cm = $10^{- 1}$ T/m

Rate of decrease of magnetic field

$\frac{d B}{d t}$ = $10^{- 3}$ T/s

Resistance of the loop, R = 4.50 mΩ = 4.5 $\times 10^{- 3}$ Ω

Rate of change of the magnetic flux due to motion of the loop in a non-uniform magnetic field is given as:

$\frac{d \emptyset}{d t}$ = A $\times \frac{d B}{d x}$ $\times v$ = $0.0144 \times 10^{- 1} \times 0.08$ = 1.152 $\times 10^{- 4}$ T $m^{2} s^{- 1}$

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

$\frac{d \emptyset'}{d t}$ = A $\times \frac{d B}{d t}$ = $0.0144 \times 10^{- 3}$ =

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6.12 Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 $\times 0.12 = 0.0144 m^{2}$

Velocity of the loop, v = 8 cm /s = 0.08 m/s

Gradient of the magnetic field along negative x – direction.

$\frac{d B}{d x}$ = $10^{- 3}$ T/cm = $10^{- 1}$ T/m

Rate of decrease of magnetic field

$\frac{d B}{d t}$ = $10^{- 3}$ T/s

Resistance of the loop, R = 4.50 mΩ = 4.5 $\times 10^{- 3}$ Ω

Rate of change of the magnetic flux due to motion of the loop in a non-uniform magnetic field is given as:

$\frac{d \emptyset}{d t}$ = A $\times \frac{d B}{d x}$ $\times v$ = $0.0144 \times 10^{- 1} \times 0.08$ = 1.152 $\times 10^{- 4}$ T $m^{2} s^{- 1}$

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

$\frac{d \emptyset'}{d t}$ = A $\times \frac{d B}{d t}$ = $0.0144 \times 10^{- 3}$ = 1.44 $\times 10^{- 5}$ T $m^{2} s^{- 1}$

Since the rate of change of the flux is the induced emf, the total emf in the loop can be calculated as:

e = 1.152 $\times 10^{- 4}$ + 1.44 $\times 10^{- 5}$ = 1.296 $\times 10^{- 4}$ V

The induced current, I = $\frac{e}{R}$ = $\frac{1.296 \times 10^{- 4}}{4.5 \times 10^{- 3}}$ = 28.8 $\times 10^{- 3}$ A

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

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