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

Concept: Magnetic field at centre of circular loop: \[ B = \frac{\mu_0 I}{2R} \] Magnetic energy density: \[ u = \frac{B^2}{2\mu_0} \] Total magnetic energy stored: \[ U = uV \] Step 1: Calculate magnetic field at centre.} \[ B = \frac{\mu_0 I}{2R} \] \[ B = \frac{4\pi\times10^{-7}\times2}{2\times(10\times10^{-2})} \] \[ B = \frac{8\pi\times10^{-7}}{0.2} \] \[ B = 4\pi\times10^{-6}\,T \]
Step 2: Find magnetic energy density.} \[ u = \frac{B^2}{2\mu_0} \] \[ u = \frac{(4\pi\times10^{-6})^2}{2\times4\pi\times10^{-7}} \]
Step 3: Find volume of cube.} \[ a = 1\,mm = 10^{-3} m \] \[ V = a^3 = (10^{-3})^3 = 10^{-9} m^3 \]
Step 4: Calculate total energy stored.} \[ U = \frac{B^2}{2\mu_0}\times V \] \[ U = \frac{(4\pi\times10^{-6})^2}{2\times4\pi\times10^{-7}}\times10^{-9} \] \[ U = 2\pi\times10^{-14}\,J \] \[ U \approx 6.28\times10^{-14}\,J \] Final Result \[ U = 6.28\times10^{-14}\,J \]