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Question: A coil of 100 turns having an average area of 100 cm² for each turn is held in a uniform field of 50...

A coil of 100 turns having an average area of 100 cm² for each turn is held in a uniform field of 50 gauss, the direction of the field being at right angle to the plane of the coil. If the field is removed in 0.01 sec, then average e.m.f induced in coil is -

A

0.5 V

B

10 V

C

20 V

D

50 V

Answer

0.5 V

Explanation

Solution

Here's how to solve this problem using Faraday's Law of Electromagnetic Induction:

  1. Given parameters:

    • Number of turns, N=100N = 100
    • Average area of each turn, A=100 cm2A = 100 \, \text{cm}^2
    • Uniform magnetic field, B=50 gaussB = 50 \, \text{gauss}
    • Time taken to remove the field, Δt=0.01 s\Delta t = 0.01 \, \text{s}

  2. Convert units to SI units:

    • Area: A=100 cm2=100×(10−2 m)2=100×10−4 m2=10−2 m2A = 100 \, \text{cm}^2 = 100 \times (10^{-2} \, \text{m})^2 = 100 \times 10^{-4} \, \text{m}^2 = 10^{-2} \, \text{m}^2
    • Magnetic field: B=50 gaussB = 50 \, \text{gauss}. Since 1 gauss=10−4 Tesla (T)1 \, \text{gauss} = 10^{-4} \, \text{Tesla (T)}, B=50×10−4 T=5×10−3 TB = 50 \times 10^{-4} \, \text{T} = 5 \times 10^{-3} \, \text{T}

  3. Initial Magnetic Flux (Φ1\Phi_1): The direction of the field is at a right angle to the plane of the coil, meaning the magnetic field lines are parallel to the area vector (normal to the coil's plane). So, the angle θ=0∘\theta = 0^\circ. The initial magnetic flux through one turn is Φ1,turn=BAcos⁡(θ)=BAcos⁡(0∘)=BA\Phi_{1, \text{turn}} = B A \cos(\theta) = B A \cos(0^\circ) = B A. The total initial magnetic flux through the coil is Φ1=NBA\Phi_1 = N B A. Φ1=100×(5×10−3 T)×(10−2 m2)\Phi_1 = 100 \times (5 \times 10^{-3} \, \text{T}) \times (10^{-2} \, \text{m}^2) Φ1=100×5×10−5 Wb=500×10−5 Wb=5×10−3 Wb\Phi_1 = 100 \times 5 \times 10^{-5} \, \text{Wb} = 500 \times 10^{-5} \, \text{Wb} = 5 \times 10^{-3} \, \text{Wb}

  4. Final Magnetic Flux (Φ2\Phi_2): The field is removed, which means the final magnetic field strength is B2=0B_2 = 0. Therefore, the final magnetic flux is Φ2=0\Phi_2 = 0.

  5. Change in Magnetic Flux (ΔΦ\Delta\Phi): ΔΦ=Φ2−Φ1=0−(5×10−3 Wb)=−5×10−3 Wb\Delta\Phi = \Phi_2 - \Phi_1 = 0 - (5 \times 10^{-3} \, \text{Wb}) = -5 \times 10^{-3} \, \text{Wb}

  6. Average Induced EMF (ε\varepsilon): According to Faraday's Law of Electromagnetic Induction, the average induced EMF is given by: ε=∣−NΔΦΔt∣\varepsilon = \left| -N \frac{\Delta\Phi}{\Delta t} \right| (We take the magnitude as EMF is typically asked for as a positive value). ε=∣−100−5×10−3 Wb0.01 s∣\varepsilon = \left| -100 \frac{-5 \times 10^{-3} \, \text{Wb}}{0.01 \, \text{s}} \right| ε=∣−100−5×10−310−2∣\varepsilon = \left| -100 \frac{-5 \times 10^{-3}}{10^{-2}} \right| ε=∣−100×(−5×10−3×102)∣\varepsilon = \left| -100 \times (-5 \times 10^{-3} \times 10^2) \right| ε=∣−100×(−5×10−1)∣\varepsilon = \left| -100 \times (-5 \times 10^{-1}) \right| ε=∣−100×(−0.5)∣\varepsilon = \left| -100 \times (-0.5) \right| ε=∣50∣\varepsilon = \left| 50 \right| ε=0.5 V\varepsilon = 0.5 \, \text{V}

The average e.m.f induced in the coil is 0.5 V.