The given physical quantity 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, and \( \mu_0 \) is the permeability of free space. We are asked to find the dimensions of \( \vec{S} \) and compare it to the dimensions of other physical quantities.
Step 1: Analyzing the dimensions of each term
The electric field \( \vec{E} \) has dimensions of force per unit charge, which can be expressed as:
\( [E] = \frac{ML}{T^3I} \)
The magnetic field \( \vec{B} \) has dimensions of force per unit charge per velocity, which can be written as:
\( [B] = \frac{M}{T^2I} \)
The permeability of free space \( \mu_0 \) has dimensions:
\( [\mu_0] = \frac{M}{A^2T^2} \)
Step 2: Calculating the dimensions of \( \vec{S} \)
The dimensions of \( \vec{S} \) are given by:
\( [\vec{S}] = \frac{[E] \times [B]}{[\mu_0]} \)
Substituting the dimensions of \( \vec{E} \), \( \vec{B} \), and \( \mu_0 \):
\( [\vec{S}] = \frac{\left( \frac{ML}{T^3I} \right) \times \left( \frac{M}{T^2I} \right)}{\frac{M}{A^2T^2}} = \frac{M^2L}{T^5I^2} \times \frac{A^2T^2}{M} = \frac{ML^2}{T^3I^2A^2} \)
Step 3: Identifying the quantity with the same dimensions
The dimensions of \( \vec{S} \) are the same as the dimensions of:
\( \frac{\text{Power}}{\text{Area}} \)
This is because power has dimensions of \( \frac{ML^2}{T^3I} \) and area has dimensions of \( L^2 \), so \( \frac{\text{Power}}{\text{Area}} \) has the same dimensions as \( \vec{S} \).
Final Answer:
\( \frac{\text{Power}}{\text{Area}} \)
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