Question

The ratio of energy stored per unit volume in a solenoid having magnetic induction B to the electrostatic energy stored per unit volume in a capacitor in electric field E is

• A. \(\frac{B^2c}{E^2}\)
• B. \(\frac{B^2c^2}{E^2}\)
• C. \(\frac{Bc^2}{E^2}\)
• D. \(\frac{B^2c^2}{E}\)
• E. \(\frac{B^2c^2}{2E^2}\)

Answer 1

The Correct Option is B

The energy stored per unit volume in a magnetic field (in a solenoid) is given by: \[ u_{\text{mag}} = \frac{B^2}{2 \mu} \] where \( B \) is the magnetic induction and \( \mu \) is the permeability of the medium (for vacuum, \( \mu = \mu_0 \)). The energy stored per unit volume in an electric field (in a capacitor) is given by: \[ u_{\text{elec}} = \frac{E^2}{2 \epsilon} \] where \( E \) is the electric field strength and \( \epsilon \) is the permittivity of the medium (for vacuum, \( \epsilon = \epsilon_0 \)). The ratio of energy stored in the magnetic field to the electric field is: \[ \frac{u_{\text{mag}}}{u_{\text{elec}}} = \frac{\frac{B^2}{2 \mu}}{\frac{E^2}{2 \epsilon}} = \frac{B^2 \epsilon}{E^2 \mu} \] In the case of free space, \( \mu = \mu_0 \) and \( \epsilon = \epsilon_0 \), and the speed of light \( c \) is related to \( \mu_0 \) and \( \epsilon_0 \) by: \[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \] Thus, the ratio becomes: \[ \frac{B^2 c^2}{E^2} \]

The correct option is (B) : \(\frac{B^2c^2}{E^2}\)

Answer 2

Given: 
Magnetic energy density in a solenoid is: \[ u_B = \frac{B^2}{2\mu_0} \] 
Electrostatic energy density in a capacitor is: \[ u_E = \frac{1}{2} \varepsilon_0 E^2 \] 
Now, taking the ratio: \[ \frac{u_B}{u_E} = \frac{\frac{B^2}{2\mu_0}}{\frac{1}{2} \varepsilon_0 E^2} = \frac{B^2}{\mu_0 \varepsilon_0 E^2} \] 
Using the relation: \[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \Rightarrow \mu_0 \varepsilon_0 = \frac{1}{c^2} \] 
Substituting: \[ \frac{u_B}{u_E} = \frac{B^2}{\left(\frac{1}{c^2}\right)E^2} = \frac{B^2 c^2}{E^2} \] 
Final Answer: \(\frac{B^2 c^2}{E^2}\)