Question

The coil of a galvanometer has $500$ turns, and each turn has an average area $3\times {10}^{-4}{\text{m}}^2$. If current of $0.5\text{A}$ pass through it and torque of $1.5\text{Nm}$ is generated in the coil. Find the strength of the magnetic field and torsional constant of coil if the angle of deflection is $\frac{\pi }{2}$.

• A. $10\text{T};9.55{\text{Nm rad}}^{-1}$
• B. $20\text{T};9.55{\text{Nm rad}}^{-1}$
• C. $10\text{T};0.955{\text{Nm rad}}^{-1}$
• D. $20\text{T};0.955{\text{Nm rad}}^{-1}$

Answer 1

The correct option is D $20\text{T};0.955{\text{Nm rad}}^{-1}$

Given:
$N=500;A=3\times {10}^{-4}{\text{m}}^2;I=0.5\text{A}$

$\tau =1.5\text{Nm};\theta =\frac{\pi }{2}$

Where,
$N=$ No. of turns
$A=$ Area of a turn
$\theta =$ Angle of deflection

Torque on a moving coil galvanometer is,

$\tau =MB\sin \theta =NIAB\sin \theta$

$\tau =500\times 0.5\times 3\times {10}^{-4}\times B\sin {90}^{\circ }$

$1.5=75\times {10}^{-3}\times B$

$\therefore B=20\text{T}$

And let $C$ is torsional constant, then

$\tau =C\theta$

$\Rightarrow C\times \frac{\pi }{2}=1.5$

$\therefore C=\frac{1.5}{\pi /2}=0.954{\text{Nm rad}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.