Question

A moving coil galvanometer experiences torque $ki$ where $i$ is the current. If $N$ coils of area $A$ and moment of inertia $I$ are kept in a magnetic field $B$.
(a) Find $k$ in terms of given parameters,
(b) If for current $i$ , the angle of deflection is $\dfrac{\pi }{2}$ , find the torsion constant of spring.
(c) If a charge Q is passed suddenly through the galvanometer find out the maximum angle of deflection.

Answer 1

Hint: In part (a) we can calculate $k$ from the expression of the relation between the torque of a galvanometer and magnetic field. In part b the torsion constant can be obtained from the relation between torque and itself. The torsion constant is equal to torque divided by the angle of deflection. To solve the c part first find out how much angular impulse is being created with the help of torque. Angular momentum is directly related to kinetic energy and kinetic energy is equal to half of the multiplication of angle of deflection and torsion constant.

Complete step by step solution:
Step 1: express the formula for the torque of a moving coil.
$\therefore \tau = NiAB$ , where $\tau $ is a torque, $N$ is the number of turns is the coil, $i$ is current, $A$ is the area of the coil, and $B$ is the magnetic field.
We are given that $\tau = ki$ . equate this equation with the above equation.
$\therefore ki = NiAB$
$ \Rightarrow k = NAB$
This is the answer to part a.
Step 2: express the formula for torsion constant in terms of torque and angle
$\therefore c = \dfrac{\tau }{\theta }$ , where $\theta $ is the angle of deflection.
Substitute the value $\dfrac{\pi }{2}$ for $\theta $ in the above expression.
$\therefore c = \dfrac{NiAB}{\dfrac{\pi}{2}}$
$ \Rightarrow c = \dfrac{2}{\pi }NiAB$
This is the answer for part b.
Step 3: Maximum angular impulse $ = \int {\tau dt} $
Therefore, maximum A.I. $ = \int {NiABdt} $
$NAB$ is not a function of time, therefore,
$\therefore A.I. = NAB\int {idt} $
We know that $\int {idt = Q} $ , therefore,
$\therefore A.I. = NABQ$ , where $Q$ is the charge passed through the galvanometer.
Step 4: The angular momentum due to the maximum angular impulse can be written as
$\therefore \int {\tau dt = I\omega } $
$ \Rightarrow I\omega = NABQ$
$ \Rightarrow \omega = \dfrac{{NABQ}}{I}$
Step 5: the rotational kinetic energy due to angular momentum
$\therefore KE = \dfrac{1}{2}I{\omega ^2}$
Substitute the value of $\omega $ from the 4th step
$\therefore KE = \dfrac{1}{2}I{\left( {\dfrac{{NABQ}}{I}} \right)^2}$
$ \Rightarrow KE = \dfrac{1}{2}\dfrac{{{N^2}{A^2}{B^2}{Q^2}}}{I}$
Step 6: Now, we know that the rotational kinetic energy can be expressed in terms of torsion constant and maximum angle of deflection.
$\therefore KE = \dfrac{1}{2}c{\theta ^2}_{\max }$
$ \Rightarrow \dfrac{1}{2}\dfrac{{{N^2}{A^2}{B^2}{Q^2}}}{I} = \dfrac{1}{2}c{\theta ^2}_{\max }$
$ \Rightarrow {\theta ^2}_{\max } = \dfrac{{{N^2}{A^2}{B^2}{Q^2}}}{{c.I}}$
Put the value of torsion constant from part b.
$\therefore {\theta ^2}_{\max } = \dfrac{{{N^2}{A^2}{B^2}{Q^2}}}{I} \times \dfrac{\pi }{{2NiAB}}$
$ \Rightarrow {\theta ^2}_{\max } = \dfrac{{\pi NAB{Q^2}}}{{2I.i}}$
$ \Rightarrow {\theta _{\max }} = Q\sqrt {\dfrac{{\pi NAB}}{{2I.i}}} $
This is the answer for part c.
Note: The angle used in torque in the first part and the first step is maximum. And we know that for the galvanometer that angle is 90 degrees. Therefore, two vector quantities will multiply but the sine of the angle between them is equal to 1. Therefore, the torque formula will not include the term sine for the vector cross multiplication.