Question

Given a charge $ q $, current $ I $ and permeability of vacuum $ \mu_0 $. Which of the following quantity has the dimension of momentum?

• A. \( qI / \mu_0 \)
• B. \( q \mu_0 I \)
• C. \( q^2 \mu_0 I \)
• D. \( q \mu_0 / I \)

Hint:

To match the dimensions of momentum, use the fact that momentum has the dimensions \( \text{M} \cdot \text{L} \cdot \text{T}^{-1} \) and check for the correct expression.

Answer 1

The Correct Option is B

We are given: \[ Q = AT \] \[ I = A \] \[ \mu_0 = \text{ML}^3 \text{T}^{-2} \text{A}^{-2} \] Now, we need to find the dimensions of the product \( P = Q \mu_0 I \). The dimension of \( P \) is calculated as follows: \[ P = Q \mu_0 I = [AT] [\text{ML}^3 \text{T}^{-2} \text{A}^{-2}] [A] \] This simplifies to: \[ P = [\text{M}^1 \text{L}^1 \text{T}^{-2} \text{A}^1] \] Now, we check the dimensions of momentum: \[ \text{Momentum} = \text{M} \cdot \text{L} \cdot \text{T}^{-1} \] We find that the dimensions of \( P \) are the same as that of momentum. Therefore, the correct answer is: \[ q \mu_0 I \]

Answer 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.