Magnetics Ques 100
- A moving coil galvanometer has a coil with $175$ turns and area $1$ $cm^{2}$. It uses a torsion band of torsion constant $10^{-6} N-m / rad$. The coil is placed in a magnetic field $B$ parallel to its plane. The coil deflects by $1^{\circ}$ for a current of $1$ $mA$. The value of $B$ (in Tesla) is approximately
(2019 Main, 9 April II)
(a) $10^{-3}$
(b) $10^{-4}$
(c) $10^{-1}$
(d) $10^{-2}$
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Answer:
Correct Answer: 100.(a)
Solution:
Formula:
- In a moving coil galvanometer in equilibrium, torque on coil due to current is balanced by torque of torsion band.
As, torque on coil,
$\tau = M \times B = NIAB$ $sin$ $\alpha$
where, $\quad B=$ magnetic field strength,
$I=$ current,
$N$ =number of turns of coil
Since, plane of the coil is parallel to the field.
$ \therefore \quad \alpha=90^{\circ} \Rightarrow \tau=N I B A $
Torque of torsion band, $T=k \theta$
where, $k=$ torsion constant of torsion band
and $\theta=$ deflection of coil in radians or angle of twist of restoring torque.
$ \therefore \quad B I N A=k \theta \text { or } B=\frac{k \theta}{I N A} $ $\quad$ …….(i)
Here,
$ \begin{aligned} k & =10^{-6} N-m / rad, \\ I & =1 \times 10^{-3} A, \\ N & =175, \\ A & =1 cm^{2}=1 \times 10^{-4} m^{2} \\ \theta & =1^{\circ}=\frac{\pi}{180} rad \end{aligned} $
Substituting values in Eq. (i), we get
$ \begin{aligned} B & =\frac{10^{-6} \times 22}{1 \times 10^{-3} \times 175 \times 7 \times 180 \times 10^{-4}} \\ & =0.998 \times 10^{-3} \simeq 10^{-3} T \end{aligned} $
BETA
