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CBSE Class 12 · 2005 · Magnetic Effects of CurrentLong Answer

(a) State Biot-Savart's law. Using this law, derive the expression for the magnetic field due to a current-carrying circular loop of radius

R

at a point at a distance

x

from its centre along the axis of the loop. (b) Two small identical circular loops (1) and (2), carrying equal currents, are placed with their geometrical axes perpendicular to each other. Find the magnitude and direction of the net magnetic field produced at the common point O. Radius of each loop is

R

.

Solution

(a) \textbf{Biot-Savart's law:}

d B = \frac{μ _{0}}{4 π} \frac{I d ℓ \times r}{r ^{3}}

. For an axial point of a circular loop, each element gives

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

; perpendicular components cancel, axial components add:

B = \frac{μ _{0} I}{4 π ( R ^{2} + x ^{2} )} \cdot \frac{R}{R ^{2} + x ^{2}} (2 π R) = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}} .

(b) Each loop gives the same axial field magnitude

B_{0} = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}

at O, directed along its own axis. The axes are perpendicular, so

B_{n e t} = B_{0}^{2} + B_{0}^{2} = 2 B_{0},

making

4 5^{\circ}

with each axis.

CBSE Class 12 Board · 2005 · Subjective

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