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Electromagnetic Induction question

2023 · 15 Apr · Shift 1 · Q70
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  5. /2023 · 15 Apr · Shift 1 · Q70

Electromagnetic Induction question

2023 · 15 Apr · Shift 1 · Q70

JEE MainPhysicsElectromagnetic InductionNumerical+4 / −1
A 20 cm20 \mathrm{~cm}20 cm long metallic rod is rotated with 210 rpm210~ \mathrm{rpm}210 rpm about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field 0.2 T0.2 \mathrm{~T}0.2 T parallel to the axis exists everywhere. The emf developed between the centre and the ring is ‾\underline{\hspace{2cm}}​mV\mathrm{mV}mV. Take π=227\pi=\frac{22}{7}π=722​
Numerical answer
View written solutionFree

Correct answer: 88

  1. Given data
  • Length of rod: l=20 cm=0.2 ml = 20\text{ cm} = 0.2\text{ m}l=20 cm=0.2 m
  • Angular speed: 210 rpm210\text{ rpm}210 rpm
  • Magnetic field: B=0.2 TB = 0.2\text{ T}B=0.2 T
  • Rod rotates about one end, with magnetic field parallel to axis of rotation.

We need the emf between the centre (axis) and the outer ring.

  1. Formula for motional emf in a rotating rod

For a rod of length lll rotating with angular velocity ω\omegaω in a uniform magnetic field BBB perpendicular to the plane of rotation, the emf between the axis and the outer end is

E=12Bωl2\mathcal{E} = \frac{1}{2} B \omega l^2E=21​Bωl2
  1. Convert rpm to rad/s
ω=210×2π60\omega = 210 \times \frac{2\pi}{60}ω=210×602π​ ω=7π rad/s\omega = 7\pi \text{ rad/s}ω=7π rad/s

Using π=227\pi = \frac{22}{7}π=722​,

ω=7×227=22 rad/s\omega = 7 \times \frac{22}{7} = 22\text{ rad/s}ω=7×722​=22 rad/s
  1. Substitute into emf formula
E=12×0.2×22×(0.2)2\mathcal{E} = \frac{1}{2} \times 0.2 \times 22 \times (0.2)^2E=21​×0.2×22×(0.2)2

First,

(0.2)2=0.04(0.2)^2 = 0.04(0.2)2=0.04

So,

E=12×0.2×22×0.04\mathcal{E} = \frac{1}{2} \times 0.2 \times 22 \times 0.04E=21​×0.2×22×0.04 E=0.1×22×0.04\mathcal{E} = 0.1 \times 22 \times 0.04E=0.1×22×0.04 E=2.2×0.04=0.088 V\mathcal{E} = 2.2 \times 0.04 = 0.088\text{ V}E=2.2×0.04=0.088 V
  1. Convert to mV
0.088 V=88 mV0.088\text{ V} = 88\text{ mV}0.088 V=88 mV
  1. Final answer

The emf developed is

88 mV\boxed{88\text{ mV}}88 mV​

This matches the stored correct answer.

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