Physics Dimensional Analysis: Momentum Calculation
This solution determines which physical quantity possesses the dimensions of momentum, given charge ($q$), current ($I$), and the permeability of vacuum (μo).
Fundamental Dimensions
- Dimension of Charge: $[q] = AT$
- Dimension of Current: $[I] = A$
- Dimension of Permeability of vacuum: $[\mu_o] = MLT^{-2}A^{-2}$
Target Dimension: Momentum
The dimension of momentum ($p$) is mass times velocity:
$[p] = M L T^{-1}$
Evaluating Option D: $qμoI$
Let's calculate the dimensions of the quantity $qμoI$:
$[q\mu_o I] = [q] \times [\mu_o] \times [I]$
Substitute the dimensions:
$[q\mu_o I] = (AT) \times (MLT^{-2}A^{-2}) \times (A)$
Combine the powers of each base dimension:
$[q\mu_o I] = M^1 L^1 T^{1-2} A^{1-2+1}$
$[q\mu_o I] = M L T^{-1} A^0 = M L T^{-1}$
Conclusion
The dimensions of $qμoI$ are $M L T^{-1}$, matching the dimensions of momentum. Thus, $qμoI$ is the correct quantity.
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Important Questions from General Physics
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