Question

-. The resistance of a moving coil voltmeter is 12,0001. The moving coil has 100
turns and is 4 cm long and 3 cm wide. The flux density in the air gap is 6 X 10-2
Wb/m2. Find the deflection produced by 300 V if the spring control gives a
deflection of one degree for a torque of 25 X 10-7 N-m.​

Answer 1

Given :

Resistance of the moving coil galvanometer ( $R_g$ ) = 120001 ohm

No of turns in the given moving coil = 100

Length of  the coil = 4 cm

Width of the coil = 3 cm

Flux density in between the air gap = $6 \times 10 ^{-2} \frac{W}{m^2}$

Torque corresponding to a deflection of one degree = $25 \times 10^7 N-m$

To Find :

Deflection produced in the coil by 300 V = /

Solution :

For the voltage of 300 V across the voltmeter , current will be ;

$I = \frac{V}{R}$

Or,$I =\frac{300}{120001} A$

So, $I$ = 0.0024 A

Due to this current  magnetic dipole moment induced in the coil will be :

= nIA

Here , n = no of  turns in the coil , I = current and A = area of the loop

And , as given magnetic flux( $\Phi$ ) = $B \times A$

For 1 $m^2$  , $\Phi$ = $6 \times 10 ^{-2} T$

Torque on this magnetic dipole :

$\vec \tau = \vec m \times \vec B$

Or,$|\vec \tau | = mB sin\theta$

Now, the deflection produced will be :

= $\frac{\tau}{25 \times 10^{-7}}$

$= \frac{mBsin \theta}{25 \times 10^{-7}}$

$= \frac{nIABsin\theta}{25 \times 10^{-7}}$

$= \frac{100 times 2.4 \times 10^{-3} \times 6 \times 10^{-2} \times 4 \times 3 \times 10^{-4}}{25 \times 10^{-7}}$

= 6.912 degree

So, the deflection produced in the coil by 300 V is 6.912 degree .