Magnetics Ques 112
- 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 $B$.
$(2005,6$ M)
(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.
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Solution:
Formula:
- Situation given in question is shown below;
Path taken by particle of charge ’ $q$ ’ and mass ’ $m$ ’ is a circle of radius $r$ where,
$ r=\frac{m v}{B q} $
Here final velocity
$ \mathbf{v} _f=\mathbf{v} _x \hat{\mathbf{i}}+\mathbf{v} _{y f}(-\hat{\mathbf{j}})=v \cos 60 \hat{\mathbf{i}}-v \sin 60^{\circ} \hat{\mathbf{j}} $
$ =v \quad (\frac{1}{2} \hat{\mathbf{i}}-\frac{\sqrt{3}}{2} \hat{\mathbf{j}}) $
So, change of velocity of charged particle is
$ \begin{aligned} \Delta \mathbf{v} & =\mathbf{v} _f-\mathbf{v} _i=v (\frac{1}{2} \hat{\mathbf{i}}-\frac{\sqrt{3}}{2} \hat{\mathbf{j}}) -v \hat{\mathbf{i}} \\ & =-v (\frac{1}{2} \hat{\mathbf{i}}+\frac{\sqrt{3}}{2} \hat{\mathbf{j}}) \end{aligned} $
It $t=$ time taken by charged particle to cross region of magnetic field then,
$ \begin{aligned} & t=\frac{\text { distance } O P}{\text { speed in direction } O P} \\ & =\frac{r \times \frac{\sqrt{3}}{2}}{v}=\frac{\frac{m v}{B q} \times \frac{\sqrt{3}}{2}}{v}=\frac{\sqrt{3} m}{2 B q} \end{aligned} $
So, acceleration of charged particle at the point its emergence is;
Acceleration,
$\mathbf{a} =\frac{\Delta \mathbf{v}}{\Delta t}=\frac{-v (\frac{1}{2} \hat{\mathbf{i}}+\frac{\sqrt{3}}{2} \hat{\mathbf{j}})}{\frac{\sqrt{3}}{2} \frac{m}{B q}} $
$ =\frac{-B q v}{m} (\frac{\hat{\mathbf{i}}}{\sqrt{3}}+\hat{\mathbf{j}}) ms^{-2}$
BETA
