Magnetic Effects & Magnetism

JEE Main level, single correct, medium

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Question 241 of 276

An insulated wire is wound so that it forms a flat coil with N=200N = 200 turns. The radius of the innermost turn is r1=3r_1 = 3 cm, and of the outermost turn r2=6r_2 = 6 cm. If 2020 mA current flows in it then the magnetic moment will be α×10−2\alpha \times 10^{-2} A.m2^2. The value of α\alpha is _____.

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Answer: (B) 2.642.64

Let the number of turns per unit radial width be n=Nr2−r1n = \dfrac{N}{r_2 - r_1}.

Consider an elemental ring of radius rr and width drdr. The number of turns in this element is dN=n drdN = n \, dr.

The magnetic moment of this elemental ring is dM=I dN A=I(Nr2−r1 dr)(πr2)dM = I \, dN \, A = I \left( \dfrac{N}{r_2 - r_1} \, dr \right) (\pi r^2).

Integrating from r1r_1 to r2r_2 to find the total magnetic moment MM:

M=∫r1r2IπNr2−r1r2 dr=IπN3(r2−r1)(r23−r13)M = \int_{r_1}^{r_2} \dfrac{I \pi N}{r_2 - r_1} r^2 \, dr = \dfrac{I \pi N}{3(r_2 - r_1)} (r_2^3 - r_1^3)

Substituting the given values: I=20×10−3I = 20 \times 10^{-3} A, N=200N = 200, r1=0.03r_1 = 0.03 m, and r2=0.06r_2 = 0.06 m:

M=20×10−3×π×2003(0.06−0.03)((0.06)3−(0.03)3)M = \dfrac{20 \times 10^{-3} \times \pi \times 200}{3(0.06 - 0.03)} \left( (0.06)^3 - (0.03)^3 \right)

M=4π0.09(216×10−6−27×10−6)M = \dfrac{4\pi}{0.09} (216 \times 10^{-6} - 27 \times 10^{-6})

M=4π0.09×189×10−6=4π×2100×10−6M = \dfrac{4\pi}{0.09} \times 189 \times 10^{-6} = 4\pi \times 2100 \times 10^{-6}

M=84π×10−4≈84×3.14×10−4=263.76×10−4≈2.64×10−2M = 84\pi \times 10^{-4} \approx 84 \times 3.14 \times 10^{-4} = 263.76 \times 10^{-4} \approx 2.64 \times 10^{-2} A.m2^2

Comparing with α×10−2\alpha \times 10^{-2} A.m2^2, we get α=2.64\alpha = 2.64.

Answer: 2.642.64

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