A current of 1 A is flowing through a very long solenoid made of winding density
3000 turns/m. As shown in the figure, a parallel plate capacitor, with plates
oriented parallel to the solenoid axis and carrying surface charge density
6\epsilon_0 C m-2, is placed at the middle of the solenoid. The momentum density of the
electromagnetic field at the midpoint X of the capacitor is n × 10-1^3 N s m-3. The
value of n is ______________ (Round off to the nearest integer)
(speed of light c = 3 × 108 m s-1)
Official Answer Key:
2 to 2
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.65)**
**1. Electrodynamics Principles (Field Momentum Density)**:
- Momentum density of an electromagnetic field in vacuum:
$\vec{g} = \epsilon_0 (\vec{E} \times \vec{B})$
- Magnetic field of long solenoid with $n_t = 3000$ turns/m and $I = 1$ A:
$B = \mu_0 n_t I$
- Electric field between capacitor plates with surface charge density $\sigma = 6\epsilon_0$:
$E = \frac{\sigma}{\epsilon_0} = \frac{6\epsilon_0}{\epsilon_0} = 6 \text{ V m}^{-1}$
**2. Detailed Step-by-Step Evaluation**:
- Since the capacitor plates are parallel to the solenoid axis, $\vec{E} \perp \vec{B}$, so $|\vec{E} \times \vec{B}| = E B$.
- Using Maxwell relation $\epsilon_0 \mu_0 = \frac{1}{c^2}$:
$g = \epsilon_0 E B = \epsilon_0 (6)(\mu_0 n_t I) = 6(\epsilon_0 \mu_0) n_t I = \frac{6 n_t I}{c^2}$
- Substitute the given numerical values:
- $n_t = 3000 \text{ m}^{-1}$
- $I = 1$ A
- $c = 3 \times 10^8 \text{ m s}^{-1} \implies c^2 = 9 \times 10^{16} \text{ m}^2\text{ s}^{-2}$
- Compute $g$:
$g = \frac{6 \times 3000 \times 1}{9 \times 10^{16}} = \frac{18000}{9 \times 10^{16}} = 2.0 \times 10^{-13} \text{ N s m}^{-3}$
- In the given format $n \times 10^{-13} \text{ N s m}^{-3}$, the integer value is $n = 2$.
**3. Official Verification**:
The verified official IIT Master Answer range is **2 to 2**.
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