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Induced Current and Induced Charge

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Estimated time: 7 minutes
CISCE: Class 12

Formula: Faraday’s Law

If a coil has N turns and the magnetic flux through it changes by ΔΦB​ in time Δt, the induced emf is:

\[{\varepsilon=-N\frac{\Delta\Phi_B}{\Delta t}}\]

CISCE: Class 12

Induced Current

If the coil circuit is closed and its total resistance is R, then the induced current is:

  • I = \[\frac {ε}{R}\]

Therefore,

  • I = \[\frac {N}{R}\frac {ΔΦ_B}{Δt}\]

Key Point

  • Induced current depends on the resistance of the circuit.
  • Induced emf is independent of circuit resistance.
CISCE: Class 12

Induced Charge

The charge flowing through the circuit in the time interval Δt is:

  • q = I Δt

Substituting the expression for induced current:

  • q = \[\frac {N}{R}\frac {ΔΦ_B}{Δt}\]× Δt
  • q = \[\frac {NΔΦ_B}{R}\]{R}}q=RNΔΦB​​​

SI Unit

If ΔΦB​ is in weber and R is in ohm, then q is obtained in coulomb.

CISCE: Class 12

Important Observations

Quantity Relation Dependence on time interval
Induced emf ε = −N\[\frac {ΔΦ_B}{Δt}\] Depends on Δt
Induced current I = \[\frac {N}{R}\frac {ΔΦ_B}{Δt}\] Depends on Δt
Induced charge q = \[\frac {NΔΦ_B}{R}\] Does not depend on Δt
 

Important: For a given change in magnetic flux, induced emf and induced current are inversely proportional to the time taken for the change.

  • ε ∝ \[\frac {1}{Δt}\], I ∝ \[\frac {1}{Δt}\]
Important: Induced charge depends only on the change in magnetic flux and resistance. It does not depend on the time taken for the flux change.
CISCE: Class 12

Example

Question: A coil of 100 turns and resistance 10 Ω encloses an area of 100 cm2. It is placed at an angle of 70° with a magnetic field of 0.1 Wb m-2. Find:

  1. Magnetic flux through the coil.
  2. Induced emf if the magnetic field is reduced to zero in 10−3 s.
  3. Charge flowing through the coil.

Given

N = 100, R = 10 Ω, A = 100 cm2 = 10−2 m2, B = 0.1 Wb m−2

The angle with the perpendicular to the plane of the coil is:

  • θ = 90° − 70° = 20°
  • cos ⁡20° = 0.94

Step 1: Magnetic flux through each turn

  • ΦB = BA cos⁡ θ = (0.1)(10−2)(0.94) = 0.94 × 10−3 Wb

Step 2: Flux through the entire coil

  • NΦB = 100 × (0.94 × 10−3) = 0.094 Wb

When the magnetic field is reduced to zero:

  • Δ(NΦB) = 0 − 0.094 = −0.094 Wb

Step 3: Induced emf

  • ε = −\[\frac {Δ(NΦ_B)}{Δt}\] = \[\frac {0.094}{10^{−3}}\] = 94 V

Step 4: Induced current

  • I = \[\frac {ε}{R}\] = \[\frac {94}{10}\] = 9.4 A

Step 5: Charge flowing through the coil

  • q = I Δt = (9.4)(10−3) = 9.4 × 10−3 C

6. Final Answers

Quantity Answer
Magnetic flux through each turn 0.94 × 10−3 Wb
Magnetic flux through the entire coil 0.094 Wb
Induced emf 94 V
Induced current 9.4 A
Charge flowing through the coil 9.4 × 10−3 C

Shaalaa.com | Electromagnetic Induction part 6 (Direction of induced current)

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Electromagnetic Induction part 6 (Direction of induced current) [00:14:01]
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Electromagnetic Induction part 6 (Direction of induced current)
00:14:01  undefined
Electromagnetic Induction part 7 (Methods of inducing EMF -1)
00:10:56  undefined
Electromagnetic Induction part 8 (Methods of inducing EMF -2)
00:11:55  undefined
Electromagnetic Induction part 9 (Numericals)
00:11:39  undefined
Electromagnetic Induction part 10 (Examples, Numericals)
00:12:03  undefined
Electromagnetic Induction part 11 (Examples, Numericals)
00:12:13  undefined
Electromagnetic Induction part 12 (Examples, Numericals)
00:04:42  undefined
Electromagnetic Induction part 16 (Back EMF)
00:06:19  undefined
Electromagnetic Induction part 21 (Examples, Numericals)
00:12:17  undefined
Electromagnetic Induction part 22 (Examples, Numericals)
00:05:53  undefined
Electromagnetic Induction part 23 (Examples, Numericals)
00:12:15  undefined
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