The dimension of B22- 0- where B is magnetic field and -0 is the magnetic permeability of vacuum- is

As we know, The quantity B^22μ _0 is the energy density of magnetic field. B^22μ _0=EV= Energy per unit volume. Therefore, [ B^22μ _0 ]= [ ML^2T^ 2L^3 ]=ML^ ...

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Standard VII

Mathematics

Area of a Circle

The dimension...

Question

The dimension of $\frac{B^{2}}{2 μ_{0}}$ , where $B$ is magnetic field and $μ_{0}$ is the magnetic permeability of vacuum, is

M L^{- 1} T^{- 2}

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M L^{2} T^{- 1}

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M L T^{- 2}

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M L^{2} T^{- 2}

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Solution

The correct option is A $M L^{- 1} T^{- 2}$
As we know,
The quantity $\frac{B^{2}}{2 μ_{0}}$ is the energy density of magnetic field.

$\frac{B^{2}}{2 μ_{0}} = \frac{E}{V}$ = Energy per unit volume.

Therefore,

$[\frac{B^{2}}{2 μ_{0}}] = [\frac{M L^{2} T^{- 2}}{L^{3}}] = M L^{- 1} T^{- 2}$

Final Answer: (𝑎)

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Area of a Circle

Standard VII Mathematics

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The dimension of B22μ0, where B is magnetic field and μ0 is the magnetic permeability of vacuum, is

The correct option is A ML−1T−2As we know, The quantity B22μ0 is the energy density of magnetic field. B22μ0=EV= Energy per unit volume. Therefore, [B22μ0]=[ML2T−2L3]=ML−1T−2 Final Answer: (𝑎)

The dimension of $\frac{B^{2}}{2 μ_{0}}$ , where $B$ is magnetic field and $μ_{0}$ is the magnetic permeability of vacuum, is