Question

The magnetic energy stored per unit volume in a solenoid with 1000 turns per metre carrying a current of 0.7 A is:

• A. \(0.154\;J\,m^{-3}\)
• B. \(0.616\;J\,m^{-3}\)
• C. \(0.308\;J\,m^{-3}\)
• D. \(0.924\;J\,m^{-3}\)

Hint:

For a solenoid, \[ u=\frac12\mu_0n^2I^2. \] This formula directly gives magnetic energy stored per unit volume.

Answer 1

The Correct Option is C

Concept: The magnetic energy density stored in a magnetic field is \[ u=\frac{B^{2}}{2\mu_{0}}. \] For a long solenoid, \[ B=\mu_0 n I. \] Combining both formulas, \[ u=\frac{(\mu_0 nI)^2}{2\mu_0} = \frac{\mu_0 n^2 I^2}{2}. \]

Step 1: Write the given quantities.
\[ n=1000\;m^{-1} \] \[ I=0.7A \] \[ \mu_0=4\pi\times10^{-7}\;H\,m^{-1} \]

Step 2: Substitute into the energy density formula.
\[ u = \frac{(4\pi\times10^{-7})(1000)^2(0.7)^2}{2} \] \[ = \frac{4\pi\times10^{-7}\times10^6\times0.49}{2} \] \[ = \frac{1.96\pi\times10^{-1}}{2} \] \[ = 0.308\;Jm^{-3} \]

Step 3: Final answer.
\[ \boxed{0.308\;Jm^{-3}} \]