Question

The formula for magnetic energy density for a solenoid is

• A. \(\frac{B^2}{2\mu_0}\)
• B. \(\frac{B}{2\mu_0}\)
• C. \(\frac{2\mu_0}{B^2}\)
• D. \(\frac{B}{4\pi\mu_0}\)

Hint:

This formula is analogous to the electric energy density in a capacitor, which is \(u_E = \frac{1}{2}\epsilon_0 E^2\). Notice the similarity: \(B^2\) corresponds to \(E^2\), and \(1/\mu_0\) corresponds to \(\epsilon_0\). This can be a useful memory aid.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A current-carrying inductor stores energy in the magnetic field within it. Magnetic energy density (\(u_B\)) is the amount of this stored energy per unit volume.
Step 2: Key Formula or Approach:
The total energy (\(U\)) stored in an inductor of inductance \(L\) carrying a current \(I\) is \(U = \frac{1}{2}LI^2\).
For an ideal long solenoid, the inductance is \(L = \mu_0 n^2 A l\) and the magnetic field inside is \(B = \mu_0 n I\), where \(n\) is the number of turns per unit length, \(A\) is the cross-sectional area, and \(l\) is the length. The volume of the solenoid is \(V = Al\).
The energy density is \(u_B = \frac{U}{V}\).
Step 3: Detailed Explanation:
From the magnetic field formula, we can express the current as \(I = \frac{B}{\mu_0 n}\).
Now substitute the expressions for \(L\) and \(I\) into the energy formula: \[ U = \frac{1}{2} (\mu_0 n^2 A l) \left(\frac{B}{\mu_0 n}\right)^2 \] \[ U = \frac{1}{2} (\mu_0 n^2 A l) \left(\frac{B^2}{\mu_0^2 n^2}\right) \] Cancel out terms (\(n^2\) and one \(\mu_0\)): \[ U = \frac{1}{2} \frac{B^2}{\mu_0} (A l) \] The energy density \(u_B\) is the energy \(U\) divided by the volume \(V = Al\): \[ u_B = \frac{U}{V} = \frac{\frac{1}{2} \frac{B^2}{\mu_0} (A l)}{A l} \] \[ u_B = \frac{B^2}{2\mu_0} \] Step 4: Final Answer:
The formula for magnetic energy density inside a solenoid (and for any magnetic field in free space) is \(\frac{B^2}{2\mu_0}\). Option (A) is correct.