CBSE Class 12 · 2020 · Magnetic Effects of CurrentShort Answer

(a) Write an expression for the magnetic moment associated with a current (

I

) carrying circular coil of radius

r

having

N

turns.
(b) Consider the above coil placed in the YZ-plane with its centre at the origin. Derive an expression for the magnetic field due to it at the point

(x, 0, 0)

Answer: (a) $m = N I π r^{2}$ .
(b) $B = \frac{μ _{0} N I r ^{2}}{2 ( r ^{2} + x ^{2} ) ^{3/2}}$ , directed along the axis (X-axis).

Solution

(a) Magnetic moment

m = N I A = N I π r^{2}

, directed along the axis. (b) By Biot-Savart, each element

I d ℓ

gives

d B = \frac{μ _{0} I d ℓ}{4 π ( r ^{2} + x ^{2} )}

perpendicular to the line joining it to P. The components perpendicular to the axis cancel by symmetry; the axial components add:

B = \int d B cos α = \frac{μ _{0} I}{4 π ( r ^{2} + x ^{2} )} \cdot \frac{r}{r ^{2} + x ^{2}} (N 2 π r)

B = \frac{μ _{0} N I r ^{2}}{2 ( r ^{2} + x ^{2} ) ^{3/2}} (along the X-axis).

CBSE Class 12 Board · 2020 · Subjective

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CBSE Class 12 · 2020 · Magnetic Effects of CurrentShort Answer

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