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Electromagnetic Induction question

2011 · Shift 0 · Q59
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Electromagnetic Induction question

2011 · Shift 0 · Q59

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A boat is moving due east in a region where the earth's magnetic fields is 5.0×10−5NA−1 m−15.0 \times {10^{ - 5}}N{A^{ - 1}}\,{m^{ - 1}}5.0×10−5NA−1m−1 due north and horizontal. The best carries a vertical aerial 2m2m2m long. If the speed of the boat is 1.50 ms−1,1.50\,m{s^{ - 1}},1.50ms−1, the magnitude of the induced emfemfemf in the wire of aerial is :
  1. A
    0.75mV0.75mV0.75mV
  2. B
    0.50mV0.50mV0.50mV
  3. C
    0.15mV0.15mV0.15mV
  4. D
    1mV1mV1mV
View written solutionFree

Correct answer: C

  1. Given data
  • Earth's magnetic field:
    B=5.0×10−5 TB = 5.0\times 10^{-5}\,\text{T}B=5.0×10−5T (since 1 T=1 N A−1m−11\,\text{T} = 1\,N\,A^{-1}m^{-1}1T=1NA−1m−1)
  • Length of vertical aerial:
    l=2 ml = 2\,\text{m}l=2m
  • Speed of boat:
    v=1.50 m s−1v = 1.50\,\text{m s}^{-1}v=1.50m s−1
  1. Direction analysis
  • Boat moves due east.
  • Magnetic field is due north and horizontal.
  • The aerial is vertical.

So, the velocity v⃗\vec vv is perpendicular to magnetic field B⃗\vec BB.

The motional emf induced across a conductor of length lll moving with speed vvv perpendicular to a magnetic field BBB is:

E=Blv\mathcal{E} = BlvE=Blv

  1. Calculation

E=(5.0×10−5)(2)(1.50)\mathcal{E} = (5.0\times 10^{-5})(2)(1.50)E=(5.0×10−5)(2)(1.50)

E=15×10−5 V\mathcal{E} = 15\times 10^{-5}\,\text{V}E=15×10−5V

E=1.5×10−4 V\mathcal{E} = 1.5\times 10^{-4}\,\text{V}E=1.5×10−4V

Convert to millivolts:

1.5×10−4 V=0.15×10−3 V=0.15 mV1.5\times 10^{-4}\,\text{V} = 0.15\times 10^{-3}\,\text{V} = 0.15\,\text{mV}1.5×10−4V=0.15×10−3V=0.15mV

  1. Option check
  • A: 0.75 mV0.75\,\text{mV}0.75mV ❌
  • B: 0.50 mV0.50\,\text{mV}0.50mV ❌
  • C: 0.15 mV0.15\,\text{mV}0.15mV ✅
  • D: 1 mV1\,\text{mV}1mV ❌

Thus, the correct option is C.

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