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

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a

ML−1T−2

b

ML2T−2

c

ML2T−1

d

MLT−2

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detailed solution

Correct option is A

Energy density in magnetic field=B22μ0=Force ×displacementdisplacement3=MLT−2LL3=ML−1T−2

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In a region of free space the electric field at some instant of time is $\overset{⇀}{E} = (80 \overset{\land}{i} + 32 \overset{\land}{j} - 64 \overset{\land}{k}) V / m$ and the magnetic field is $\overset{⇀}{B} = (0.2 \overset{\land}{i} + 0.08 \overset{\land}{j} - 0.29 \overset{\land}{k}) μ T$ . The pointing vector for these fields is

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