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JEE Main 2020 (7 Jan, Shift 2) · Units & Dimensions

Units & Dimensions question from JEE Main 2020 (7 Jan, Shift 2)

Q73JEE Main 2020 · 7 JAN · SHIFT 2Easy

Units & Dimensions

The dimension of B22μ0\frac{B^2}{2\mu_0}, where B is magnetic field and μ0\mu_0 is the magnetic permeability of vacuum, is:

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Answer: (c)

  1. B2/(2μ0)B^2/(2\mu_0) is the energy stored per unit volume in a magnetic field.
  2. [energy/volume]=ML2T−2/L3=ML−1T−2[\text{energy}/\text{volume}] = \mathrm{ML^2T^{-2}}/\mathrm{L^3} = \mathrm{ML^{-1}T^{-2}}
  3. So the answer is (c).
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