CBSE Class 12 · 2005 · Magnetic Effects of CurrentLong Answer

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(a) With the help of a labelled diagram, explain the principle and working of a moving coil galvanometer.
(b) Two parallel coaxial circular coils of equal radius

R

and equal number of turns

N

carry equal currents

I

in the same direction and are separated by a distance

2 R

. Find the magnitude and direction of the net magnetic field produced at the mid-point of the line joining their centres.

Solution

(a) \textbf{Principle:} a current-carrying coil in a magnetic field experiences a torque. The coil, wound on a soft-iron core and hung in the radial field of a horseshoe magnet, feels a deflecting torque

τ = N I A B

; the suspension provides a restoring torque

k ϕ

. At balance

N I A B = k ϕ

, so

ϕ = \frac{N A B}{k} I \propto I

— a linear current scale. (b) Each coil is at axial distance

R

from the midpoint, so each produces

B_{0} = \frac{μ _{0} N I R ^{2}}{2 ( R ^{2} + R ^{2} ) ^{3/2}} = \frac{μ _{0} N I R ^{2}}{2 ( 2 R ^{2} ) ^{3/2}} = \frac{μ _{0} N I}{2 ^{5/2} R} .

Both point the same way (currents in the same sense), so

B = 2 B_{0} = \frac{μ _{0} N I}{2 ^{3/2} R} = \frac{μ _{0} N I}{2 2 R},

directed along the common axis.

CBSE Class 12 Board · 2005 · Subjective

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CBSE Class 12 · 2005 · Magnetic Effects of CurrentLong Answer

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