Question

A cube of side \(1\,\text{mm}\) is placed at centre of circular coil of radius \(10\,\text{cm}\). Current flowing in coil is \(2\,A\). Find magnetic field energy stored in cube. (\(\pi = 3.14\))

• A. \(3.17 \times 10^{-14}\,J\)
• B. \(6.28 \times 10^{-14}\,J\)
• C. \(12.56 \times 10^{-14}\,J\)
• D. \(9.42 \times 10^{-14}\,J\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Because the cube (\(1 \, \text{mm}\)) is negligibly small compared to the coil's radius (\(10 \, \text{cm}\)), we can approximate the magnetic field inside the entire volume of the cube as uniform and equal to the field exactly at the center of the coil.
Step 2: Key Formula or Approach:
Magnetic field at the center of a circular coil: \(B = \frac{\mu_0 I}{2R}\)
Magnetic energy density: \(u = \frac{B^2}{2\mu_0}\)
Total energy stored: \(U = u \times \text{Volume}\)
Step 3: Detailed Explanation:
First, calculate the magnetic field \(B\): \[ B = \frac{\mu_0 I}{2R} = \frac{(4\pi \times 10^{-7}) \times 2}{2 \times 0.1} \] \[ B = \frac{4\pi \times 10^{-7} \times 2}{0.2} = 4\pi \times 10^{-6} \, \text{T} \] Now, calculate the stored energy \(U\). The volume of the cube is \(V = (10^{-3})^3 = 10^{-9} \, \text{m}^3\). \[ U = \frac{B^2}{2\mu_0} \times V \] \[ U = \frac{(4\pi \times 10^{-6})^2}{2 \times (4\pi \times 10^{-7})} \times 10^{-9} \] \[ U = \frac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} \times 10^{-9} \] \[ U = (2\pi \times 10^{-5}) \times 10^{-9} = 2\pi \times 10^{-14} \, \text{J} \] Using \(\pi = 3.14\): \[ U = 2 \times 3.14 \times 10^{-14} = 6.28 \times 10^{-14} \, \text{J} \]
Step 4: Final Answer:
The energy stored in the cube is \(6.28 \times 10^{-14} \, \text{J}\).