A physical quantity $\vec{ S }$ is defined as \(\vec{ S }=\frac{(\vec{ E } \times \vec{ B }) }{ \mu_{0}}\), where $\vec{ E }$ is electric field, $\vec{ B }$ is magnetic field and $\mu_{0}$ is the permeability of free space. The dimensions of $\vec{ S }$ are the same as the dimensions of which of the following quantity(ies)?
- $\frac{\text { Energy }}{\text { Charge } \times \text { Current }}$
- $\frac{\text { Force }}{\text { Lenght } \times \text { Time }}$
- $\frac{\text { Energy }}{\text { Volume }}$
- $\frac{\text { Power }}{\text { Area }}$
The Correct Option is B, D
Solution and Explanation
Step 1: Given Information
The physical quantity \( \vec{S} \) is defined as:
\[
\vec{S} = \frac{(\vec{E} \times \vec{B})}{\mu_0}
\]
where:
- \( \vec{E} \) is the electric field,
- \( \vec{B} \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space.
We are asked to determine the dimensions of \( \vec{S} \) and match them with the dimensions of one of the given quantities.
Step 2: Determining the Dimensions of \( \vec{S} \)
The dimensions of \( \vec{S} \) depend on the cross product of the electric field \( \vec{E} \) and the magnetic field \( \vec{B} \), and the permeability of free space \( \mu_0 \). We start by analyzing the dimensions of each term:
- The dimensions of the electric field \( \vec{E} \) are given by:
\[
[E] = \frac{M L^2}{T^3 A}
\]
where \( M \) is mass, \( L \) is length, \( T \) is time, and \( A \) is the electric current.
- The dimensions of the magnetic field \( \vec{B} \) are:
\[
[B] = \frac{M}{T^2 A}
\]
- The permeability of free space \( \mu_0 \) has the dimensions:
\[
[\mu_0] = \frac{M}{A^2 T^2}
\]
Now, the cross product \( \vec{E} \times \vec{B} \) results in the dimensions:
\[
[\vec{E} \times \vec{B}] = \left[\frac{M L^2}{T^3 A} \times \frac{M}{T^2 A}\right] = \frac{M^2 L^2}{T^5 A^2}
\]
Dividing by \( \mu_0 \), we get the dimensions of \( \vec{S} \):
\[
[\vec{S}] = \frac{\frac{M^2 L^2}{T^5 A^2}}{\frac{M}{A^2 T^2}} = \frac{M L^2}{T^3 A}
\]
Step 3: Matching the Dimensions
Now, we compare the dimensions of \( \vec{S} \) with the given options. We are asked to find the quantity that has the same dimensions as \( \vec{S} \). The options are:
- (B) \( \frac{\text{Force}}{\text{Length} \times \text{Time}} \)
- (D) \( \frac{\text{Force}}{\text{Length} \times \text{Time}} \)
The dimensions of force \( F \) are:
\[
[F] = \frac{M L}{T^2}
\]
Therefore, the dimensions of \( \frac{\text{Force}}{\text{Length} \times \text{Time}} \) are:
\[
\left[\frac{F}{L T}\right] = \frac{M L}{T^2} \times \frac{1}{L T} = \frac{M}{T^3 A}
\]
This matches the dimensions of \( \vec{S} \). Therefore, both options (B) and (D) are correct.
Final Answer:
The correct options are:
- (B) \( \frac{\text{Force}}{\text{Length} \times \text{Time}} \)
- (D) \( \frac{\text{Force}}{\text{Length} \times \text{Time}} \)
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