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Units and Measurements question

2025 · 22 Jan · Shift 1 · Q51
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Units and Measurements question

2025 · 22 Jan · Shift 1 · Q51

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
If BBB is magnetic field and μ0\mu_0μ0​ is permeability of free space, then the dimensions of (B/μ0)\left(B / \mu_0\right)(B/μ0​) is
  1. A
    MT−2 A−1\mathrm{MT}^{-2} \mathrm{~A}^{-1}MT−2 A−1
  2. B
    L−1 A\mathrm{L}^{-1} \mathrm{~A}L−1 A
  3. C
    ML2 T−2 A−1\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}ML2 T−2 A−1
  4. D
    LT−2 A−1\mathrm{LT}^{-2} \mathrm{~A}^{-1}LT−2 A−1
View written solutionFree

Correct answer: B

  1. Write dimensions of magnetic field BBB

Using the magnetic force relation on a current-carrying conductor, F=BIlF = B I lF=BIl So, B=FIlB = \frac{F}{Il}B=IlF​

Now, [F]=MLT−2,[I]=A,[l]=L[F] = MLT^{-2}, \quad [I]=A, \quad [l]=L[F]=MLT−2,[I]=A,[l]=L

Hence, [B]=MLT−2AL=MT−2A−1[B] = \frac{MLT^{-2}}{AL} = MT^{-2}A^{-1}[B]=ALMLT−2​=MT−2A−1

  1. Write dimensions of permeability of free space μ0\mu_0μ0​

From the force per unit length between two parallel current-carrying wires, Fl=μ0I1I22πd\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}lF​=2πdμ0​I1​I2​​

Thus, μ0=(F/l) dI2\mu_0 = \frac{(F/l)\, d}{I^2}μ0​=I2(F/l)d​

Now, [Fl]=MLT−2L=MT−2\left[\frac{F}{l}\right] = \frac{MLT^{-2}}{L} = MT^{-2}[lF​]=LMLT−2​=MT−2

Therefore, [μ0]=(MT−2)(L)A2=MLT−2A−2[\mu_0] = \frac{(MT^{-2})(L)}{A^2} = MLT^{-2}A^{-2}[μ0​]=A2(MT−2)(L)​=MLT−2A−2

  1. Find dimensions of (B/μ0)\left(B/\mu_0\right)(B/μ0​)

[Bμ0]=MT−2A−1MLT−2A−2\left[\frac{B}{\mu_0}\right] = \frac{MT^{-2}A^{-1}}{MLT^{-2}A^{-2}}[μ0​B​]=MLT−2A−2MT−2A−1​

Subtracting powers, [Bμ0]=L−1A\left[\frac{B}{\mu_0}\right] = L^{-1}A[μ0​B​]=L−1A

  1. Match with the options

L−1AL^{-1}AL−1A

This matches Option B.

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