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Units and Measurements question

2008 · Shift 0 · Q88
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Units and Measurements question

2008 · Shift 0 · Q88

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The dimension of magnetic field in M, L, T and C (coulomb) is given as
  1. A
    MLT-1C-1
  2. B
    MT2C-2
  3. C
    MT-1C-1
  4. D
    MT-2C-1
View written solutionFree

Correct answer: C

  1. We use the magnetic force on a moving charge:

F=qvBsin⁡θF = q v B \sin\thetaF=qvBsinθ

For dimensions, take sin⁡θ\sin\thetasinθ as dimensionless, so

B=FqvB = \frac{F}{qv}B=qvF​

  1. Write dimensions of each quantity:
  • Force: [F]=MLT−2[F] = MLT^{-2}[F]=MLT−2
  • Charge: [q]=C[q] = C[q]=C
  • Velocity: [v]=LT−1[v] = LT^{-1}[v]=LT−1
  1. Substitute into the expression for BBB:

[B]=MLT−2C⋅LT−1[B] = \frac{MLT^{-2}}{C\cdot LT^{-1}}[B]=C⋅LT−1MLT−2​

  1. Simplify:

[B]=ML1−1T−2−(−1)C−1=MT−1C−1[B] = M L^{1-1} T^{-2-(-1)} C^{-1} = MT^{-1}C^{-1}[B]=ML1−1T−2−(−1)C−1=MT−1C−1

So, the dimension of magnetic field is

MT−1C−1\boxed{MT^{-1}C^{-1}}MT−1C−1​

  1. Match with options:
  • A: MLT−1C−1MLT^{-1}C^{-1}MLT−1C−1
  • B: MT2C−2MT^{2}C^{-2}MT2C−2
  • C: MT−1C−1MT^{-1}C^{-1}MT−1C−1
  • D: MT−2C−1MT^{-2}C^{-1}MT−2C−1

Hence, the correct option is:

C\boxed{\text{C}}C​

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