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

6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

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Payal Gupta | Contributor-Level 10

4 months ago

6.3 Number of turns on the solenoid = 15 turns per cm = 1500 turns per m

Hence, number of turns per unit length, n = 1500

The loop area in the solenoid, A = 2.0 ${c m}^{2}$ = 2 $\times 10^{- 4}$ $m^{2}$

Current carrying by the solenoid, changes from 2 .0 A to 4.0 A

So, change of current, di = 4 – 2 = 2 A

Change in time, dt = 0.1 s

According to Faraday’s law, the induced emf, e = $\frac{d φ}{d t}$ …………. (i)

Where $φ$ = induced flux through the small loop = BA …………. (ii)

Magnetic field is given by B = $μ_{0} n i$ …………………………&h

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6.3 Number of turns on the solenoid = 15 turns per cm = 1500 turns per m

Hence, number of turns per unit length, n = 1500

The loop area in the solenoid, A = 2.0 ${c m}^{2}$ = 2 $\times 10^{- 4}$ $m^{2}$

Current carrying by the solenoid, changes from 2 .0 A to 4.0 A

So, change of current, di = 4 – 2 = 2 A

Change in time, dt = 0.1 s

According to Faraday’s law, the induced emf, e = $\frac{d φ}{d t}$ …………. (i)

Where $φ$ = induced flux through the small loop = BA …………. (ii)

Magnetic field is given by B = $μ_{0} n i$ ……………………………. (iii)

Where $μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ H/m

From equation (i),

e = $\frac{d}{d t} (B A)$

= A $μ_{0} n \times \frac{d i}{d t}$

= 2 $\times 10^{- 4} \times$ 4 $π \times 10^{- 7} \times 1500 \times \frac{2}{0.1}$

= 7.54 $\times 10^{- 6}$ V

Hence the induced emf while current is changing is 7.54 $\times 10^{- 6}$ V

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6.3 Number of turns on the solenoid = 15 turns per cm = 1500 turns per mHence, number of turns per unit length, n = 1500The loop area in the solenoid, A = 2.0 <math><msup><mrow><mrow><mi>c</mi><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 2 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup></math> <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math>Current carrying by the solenoid, changes from 2 .0 A to 4.0 ASo, change of current, di = 4 – 2 = 2 AChange in time, dt = 0.1 sAccording to Faraday’s law, the induced emf, e = <math><mfrac><mrow><mrow><mi>d</mi><mi>φ</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>t</mi></mrow></mrow></mfrac></math> …………. (i)Where <math><mi>φ</mi></math> = induced flux through the small loop = BA …………. (ii)Magnetic field is given by B = <math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>n</mi><mi>i</mi></math> ……………………………. (iii)Where <math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math> = Permeability of free space = 4 <math><mi>π</mi><mi></mi><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup></math> H/mFrom equation (i), e = <math><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>t</mi></mrow></mrow></mfrac><mo> (</mo><mi>B</mi><mi>A</mi><mo>)</mo></math>= A <math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>n</mi><mo>×</mo><mfrac><mrow><mrow><mi>d</mi><mi>i</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>t</mi></mrow></mrow></mfrac></math>= 2 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup><mo>×</mo></math> 4 <math><mi>π</mi><mi></mi><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup><mo>×</mo><mn>1500</mn><mo>×</mo><mfrac><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>0.1</mn></mrow></mrow></mfrac></math>= 7.54 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup></math> VHence the induced emf while current is changing is 7.54 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup></math> V

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6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

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6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

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6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?