Magnetic Effects & Magnetism

JEE Main level, single correct, medium

17 of 276

Question 17 of 276

A small cube of side 11 mm is placed at the centre of a circular loop of radius 1010 cm carrying a current of 22 A. The magnetic energy stored inside the cube is α×10−14\alpha \times 10^{-14} J. The value of α\alpha is _______. (μo=4π×10−7\mu_o = 4\pi \times 10^{-7} Tm/A, π=3.14\pi = 3.14)

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Answer: (A) 6.286.28

The magnetic field at the centre of the circular loop is given by B=μ0I2RB = \dfrac{\mu_0 I}{2R}

Substituting the given values, B=4π×10−7×22×0.1=4π×10−6B = \dfrac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6} T

The magnetic energy density is u=B22μ0u = \dfrac{B^2}{2\mu_0}

u=(4π×10−6)22×4π×10−7=16π2×10−128π×10−7=2π×10−5u = \dfrac{(4\pi \times 10^{-6})^2}{2 \times 4\pi \times 10^{-7}} = \dfrac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5} J/m3^3

The volume of the small cube is V=(1 mm)3=(10−3 m)3=10−9V = (1 \text{ mm})^3 = (10^{-3} \text{ m})^3 = 10^{-9} m3^3

The magnetic energy stored inside the cube is U=u×VU = u \times V

U=2π×10−5×10−9=2π×10−14U = 2\pi \times 10^{-5} \times 10^{-9} = 2\pi \times 10^{-14} J

Using π=3.14\pi = 3.14, we get U=2×3.14×10−14=6.28×10−14U = 2 \times 3.14 \times 10^{-14} = 6.28 \times 10^{-14} J

Comparing with α×10−14\alpha \times 10^{-14} J, we get α=6.28\alpha = 6.28

Answer: 6.286.28

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