Given a charge $ q $, current $ I $ and permeability of vacuum $ \mu_0 $. Which of the following quantity has the dimension of momentum?
Show Hint
- \( qI / \mu_0 \)
- \( q \mu_0 I \)
- \( q^2 \mu_0 I \)
- \( q \mu_0 / I \)
The Correct Option is B
Approach Solution - 1
Approach Solution -2
We are given physical quantities: charge \( q \), current \( I \), and permeability of free space \( \mu_0 \). We must determine which combination of these quantities has the dimension of momentum.
Concept Used:
The dimension of momentum is given by:
\[ [\text{Momentum}] = [M\,L\,T^{-1}] \]
We must form a combination of \( q, I, \mu_0 \) that results in the same dimension.
Step-by-Step Solution:
Step 1: Write the dimensional formula of each quantity.
For charge \( q \):
\[ [q] = [I \times T] = [A\,T] \]
For current \( I \):
\[ [I] = [A] \]
For permeability of vacuum \( \mu_0 \):
\[ [\mu_0] = [M\,L\,T^{-2}\,A^{-2}] \]
Step 2: Assume the required combination is \( q^a I^b \mu_0^c \) and equate its dimensions to momentum.
\[ [A^a T^a] [A^b] [M^c L^c T^{-2c} A^{-2c}] = [M^1 L^1 T^{-1}] \]
Combine exponents for each fundamental dimension:
\[ [M]:\ c = 1 \] \[ [L]:\ c = 1 \] \[ [T]:\ a - 2c = -1 \] \[ [A]:\ a + b - 2c = 0 \]
Step 3: Substitute \( c = 1 \) into the other equations:
\[ a - 2(1) = -1 \Rightarrow a = 1 \] \[ a + b - 2(1) = 0 \Rightarrow 1 + b - 2 = 0 \Rightarrow b = 1 \]
Step 4: Therefore, the required combination is:
\[ q^1 I^1 \mu_0^1 = q I \mu_0 \]
Final Computation & Result:
The quantity having the dimension of momentum is:
\[ \boxed{\mu_0 q I} \]
Hence, \( \mu_0 q I \) has the same dimensions as momentum.
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