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(a) In a moving coil galvanometer, why is the magnetic field required to be radial?
(b) A 100 turn closely wound circular coil of radius 10 cm carries a current of 3·2 A. Calculate (i) the magnetic field at the centre of the coil, and (b) its magnetic moment.

Answers (1)

a) In moving-coil galvanometer uniform radial magnetic field is used which help to get maximum torque.

We know the equation of torque is $\tau =NIAB$

When the magnetic field is radial the angle between B and A is $90^{\circ}$ . $\sin90=1$ . So we get maximum torque.

b) Given :

radius $r=10\: cm=0.1\: m$

Number of turns $N=100$

Current $i=3.2A$

i) The expression for magnetic field is ,

$B=\left ( \frac{\mu _0}{4\pi } \right )\left ( \frac{2\pi Ni}{r} \right )$

$B=\left ( \frac{\mu _0Ni}{2r } \right )$

$=\frac{4\pi \times 10^{-7}\times 100\times 3.2}{2\times 0.1}$

$=2\times 10^{-3}T$

ii) Magnetic moment :

$m=NiA$

$=Ni\pi r^2$

$=100\times 3.2\times 3.14\times (0.1)^2$

$=10Am^2$

Thus the magnetic field is $2\times 10^{-3}T$ and magnetic moment is $10Am^2$

Posted by

Safeer PP

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