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Electromagnetic Induction question

2020 · 2 Sep · Shift 1 · Q57
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Electromagnetic Induction question

2020 · 2 Sep · Shift 1 · Q57

JEE MainPhysicsElectromagnetic InductionNumerical+4 / −1
A circular coil of radius 10 cm is placed in a uniform magnetic field of 3.0 ×\times× 10–5 T with its plane perpendicular to the field initially. It is rotated at constant angular speed about an axis along the diameter of coil and perpendicular to magnetic field so that it undergoes half of rotation in 0.2 s. The maximum value of EMF induced (in μ\muμ V) in the coil will be close to the integer ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 15

  1. Given data
  • Radius of coil: r=10 cm=0.1 mr = 10\text{ cm} = 0.1\text{ m}r=10 cm=0.1 m
  • Magnetic field: B=3.0×10−5 TB = 3.0 \times 10^{-5}\text{ T}B=3.0×10−5 T
  • Plane of coil is initially perpendicular to the field.
  • Coil rotates about a diameter, and this axis is perpendicular to B⃗\vec BB.
  • Time for half rotation: t=0.2 st = 0.2\text{ s}t=0.2 s

We assume the coil has one turn, since number of turns is not given.

  1. Magnetic flux through the coil

Magnetic flux is

Φ=BAcos⁡θ\Phi = BA\cos\thetaΦ=BAcosθ

where θ\thetaθ is the angle between the magnetic field and the normal to the coil.

Initially, the plane is perpendicular to the field, so the normal is along the field, hence flux is maximum.

As the coil rotates with angular speed ω\omegaω, the flux varies as

Φ=BAcos⁡(ωt)\Phi = BA\cos(\omega t)Φ=BAcos(ωt)

So induced emf is

e=∣−dΦdt∣=BAω ∣sin⁡(ωt)∣e = \left| -\frac{d\Phi}{dt} \right| = BA\omega\,|\sin(\omega t)|e=​−dtdΦ​​=BAω∣sin(ωt)∣

Therefore, the maximum emf is

emax⁡=BAωe_{\max} = BA\omegaemax​=BAω
  1. Find angular speed

Half rotation means angular displacement

Δθ=π\Delta \theta = \piΔθ=π

Given this occurs in 0.2 s0.2\text{ s}0.2 s,

ω=π0.2=5π rad/s\omega = \frac{\pi}{0.2} = 5\pi\ \text{rad/s}ω=0.2π​=5π rad/s
  1. Area of the coil
A=πr2=π(0.1)2=0.01π m2A = \pi r^2 = \pi (0.1)^2 = 0.01\pi\ \text{m}^2A=πr2=π(0.1)2=0.01π m2
  1. Calculate maximum emf
emax⁡=BAωe_{\max} = BA\omegaemax​=BAω

Substitute values:

emax⁡=(3.0×10−5)(0.01π)(5π)e_{\max} = (3.0\times 10^{-5})(0.01\pi)(5\pi)emax​=(3.0×10−5)(0.01π)(5π) emax⁡=3.0×10−7⋅5π2e_{\max} = 3.0\times 10^{-7} \cdot 5\pi^2emax​=3.0×10−7⋅5π2 emax⁡=15π2×10−7e_{\max} = 15\pi^2 \times 10^{-7}emax​=15π2×10−7

Using π2≈9.87\pi^2 \approx 9.87π2≈9.87,

emax⁡≈15×9.87×10−7e_{\max} \approx 15 \times 9.87 \times 10^{-7}emax​≈15×9.87×10−7 emax⁡≈148×10−7e_{\max} \approx 148 \times 10^{-7}emax​≈148×10−7 emax⁡≈1.48×10−5 Ve_{\max} \approx 1.48 \times 10^{-5}\text{ V}emax​≈1.48×10−5 V

Convert to μ\muμV:

1.48×10−5 V=14.8 μV1.48\times 10^{-5}\text{ V} = 14.8\,\mu\text{V}1.48×10−5 V=14.8μV

So the closest integer is

15\boxed{15}15​
  1. Comparison with stored answer

Stored correct answer = 151515.

Our derived answer matches it.

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