Magnetics Ques 3
- A moving coil galvanometer experiences torque $=k i$, where $i$ is current. If $N$ coils of area $A$ each and moment of inertia $I$ is kept in magnetic field $B$.
(2005, 6M)
(a) Find $k$ in terms of given parameters.
(b) If for current $i$ deflection is $\frac{\pi}{2}$, find out torsional constant of spring.
(c) If a charge $Q$ is passed suddenly through the galvanometer, find out maximum angle of deflection.
(c) $R_A R_V=G^2\left(\frac{I_g}{I_0-I_g}\right)$ and $\frac{R_A}{R_V}=\left(\frac{I_0-I_g}{I_g}\right)^2$
(d) $R_A R_V=G^2$ and $\frac{R_A}{R_V}=\frac{I_g}{\left(I_0-I_g\right)}$
Show Answer
Answer:
Correct Answer: 3. (a) $k=B N A$ (b) $k=\frac{2 B i N A}{\pi}$ (c) $Q \sqrt{\frac{B N A \pi}{2 I}}$
Solution:
- $\tau=M B=k i \Rightarrow k=\frac{M B}{i}=\frac{(N i A) B}{i}=N B A$
(b) $\tau=k \cdot \theta=B i N A$
$\therefore \quad k=\frac{2 B i N A}{\pi}$ (as $\theta=\pi / 2$ )
(c) $\tau=B i N A$
or
$ \begin{aligned} \int_0^t \tau d t & =B N A \int_0^t i d t \\ I \omega & =B N A Q \\ \omega & =\frac{B N A Q}{I} \end{aligned} $
or
At maximum deflection, whole kinetic energy (rotational) will be converted into potential energy of spring.
Hence, $\quad \frac{1}{2} I \omega^2=\frac{1}{2} k \theta_{\max }^2$
Substituting the values, we get
$ \theta_{\max }=Q \sqrt{\frac{B N \pi A}{2 I}} $
BETA
