Match the following
`{:("List-I","List-II"),(P."magnetic field",1.[AL^(2)]),(Q."Magnetic moment",2.[ATN^(-1)]),(R."Ratio of magnetic moment to angular momentum",3. [MA^(-1)T^(-2)]),(S. sqrt(epsilon_(0) mu_(0)),4.[L^(-1) T]):}`

Options

A. 3,1,2,4
B. 2,1,3,4
C. 4,3,2,1
D. 4,1,2,3

Answer

To solve the matching question regarding magnetic concepts, we need to determine the dimensions of each item in List-I and match them with the corresponding items in List-II.

### Step-by-Step Solution:

1. **Magnetic Field (P)**:
- The magnetic field (B) can be derived from the relationship between electric field (E), force (F), and charge (q). The formula is given as:
\[ B = \frac{F}{q \cdot c} \]
- The dimensions of force (F) are \( [M L T^{-2}] \) and charge (q) has dimensions of \( [A T] \) (where A is Ampere).
- The speed of light (c) has dimensions of \( [L T^{-1}] \).
- Substituting these into the formula:
\[ B = \frac{[M L T^{-2}]}{[A T] \cdot [L T^{-1}]} = \frac{[M L T^{-2}]}{[A L T^{-1}]} = [M A^{-1} T^{-2}] \]
- Thus, the dimension of magnetic field (B) is \( [M A^{-1} T^{-2}] \). This matches with option 3.

2. **Magnetic Moment (Q)**:
- The magnetic moment (μ) is defined as the product of current (I) and area (A):
\[ \mu = I \cdot A \]
- The dimensions of current (I) are \( [A] \) and area (A) is \( [L^2] \).
- Therefore, the dimensions of magnetic moment are:
\[ [\mu] = [A] \cdot [L^2] = [A L^2] \]
- This matches with option 1.

3. **Ratio of Magnetic Moment to Angular Momentum (R)**:
- Angular momentum (L) is given by the product of mass (M), velocity (V), and radius (R):
\[ L = M \cdot V \cdot R \]
- The dimensions of angular momentum are:
\[ [L] = [M] \cdot [L T^{-1}] \cdot [L] = [M L^2 T^{-1}] \]
- The ratio of magnetic moment to angular momentum is:
\[ \frac{\mu}{L} = \frac{[A L^2]}{[M L^2 T^{-1}]} = [A M^{-1} T] \]
- This matches with option 2.

4. **Square Root of (ε₀μ₀) (S)**:
- The speed of light (c) is related to ε₀ and μ₀ by the equation:
\[ c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} \]
- The dimensions of ε₀ and μ₀ can be derived from the above relationship. The dimensions of c are \( [L T^{-1}] \), thus:
\[ \sqrt{\epsilon_0 \mu_0} = \frac{1}{c} \implies [\sqrt{\epsilon_0 \mu_0}] = [T L^{-1}] \]
- This matches with option 4.

### Final Matching:
- P matches with 3.
- Q matches with 1.
- R matches with 2.
- S matches with 4.