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Question: A small cube of volume 1 $mm^3$ placed at the centre of circular loop of radius of 100 cm, carrying ...

A small cube of volume 1 mm3mm^3 placed at the centre of circular loop of radius of 100 cm, carrying a current 1A. The magnetic energy stored in side cube is :-

A

π2×1016J\frac{\pi}{2} \times 10^{16} J

B

2π×10−7J2\pi \times 10^{-7} J

C

π2×10−16J\frac{\pi}{2} \times 10^{-16} J

D

π2×10−7J\frac{\pi}{2} \times 10^{-7} J

Answer

(3)

Explanation

Solution

  1. Calculate the magnetic field at the center of the circular loop:

The magnetic field (BB) at the center of a circular loop of radius RR carrying current II is given by the formula:

B=μ0I2RB = \frac{\mu_0 I}{2R}

Given:

Current I=1 AI = 1 \, A

Radius R=100 cm=1 mR = 100 \, cm = 1 \, m

Permeability of free space μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A

Substitute the values:

B=(4π×10−7 T⋅m/A)×(1 A)2×(1 m)B = \frac{(4\pi \times 10^{-7} \, T \cdot m/A) \times (1 \, A)}{2 \times (1 \, m)}

B=2π×10−7 TB = 2\pi \times 10^{-7} \, T

  1. Calculate the magnetic energy density:

The magnetic energy density (uBu_B) in a region with magnetic field BB is given by:

uB=B22μ0u_B = \frac{B^2}{2\mu_0}

Substitute the calculated BB and μ0\mu_0:

uB=(2π×10−7 T)22×(4π×10−7 T⋅m/A)u_B = \frac{(2\pi \times 10^{-7} \, T)^2}{2 \times (4\pi \times 10^{-7} \, T \cdot m/A)}

uB=4π2×10−148π×10−7 J/m3u_B = \frac{4\pi^2 \times 10^{-14}}{8\pi \times 10^{-7}} \, J/m^3

uB=π×10−142×10−7 J/m3u_B = \frac{\pi \times 10^{-14}}{2 \times 10^{-7}} \, J/m^3

uB=π2×10−7 J/m3u_B = \frac{\pi}{2} \times 10^{-7} \, J/m^3

  1. Calculate the total magnetic energy stored in the cube:

The volume of the cube is V=1 mm3V = 1 \, mm^3. Convert this to cubic meters:

V=1 mm3=(1×10−3 m)3=1×10−9 m3V = 1 \, mm^3 = (1 \times 10^{-3} \, m)^3 = 1 \times 10^{-9} \, m^3

Since the cube is very small and placed at the center, we can assume the magnetic field is uniform throughout its volume. The total magnetic energy (UBU_B) stored in the cube is the product of energy density and volume:

UB=uB×VU_B = u_B \times V

UB=(π2×10−7 J/m3)×(1×10−9 m3)U_B = \left(\frac{\pi}{2} \times 10^{-7} \, J/m^3\right) \times (1 \times 10^{-9} \, m^3)

UB=π2×10−7×10−9 JU_B = \frac{\pi}{2} \times 10^{-7} \times 10^{-9} \, J

UB=π2×10−16 JU_B = \frac{\pi}{2} \times 10^{-16} \, J

Comparing this result with the given options, option (3) matches our calculated value.