A moving coil galvanometer has a coil with 17 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The torque on the coil in a magnetic field is given by $	au = N I A B \sin(	heta)$. Since the magnetic field is parallel to the plane of the coi

Vedclass · Accepted Answer

$10^{-2}$

Vedclass · Answer

(D) The torque on the coil in a magnetic field is given by $	au = N I A B \sin(	heta)$. Since the magnetic field is parallel to the plane of the coil,the angle between the normal to the coil and the magnetic field is $90^{\circ}$,so $\sin(90^{\circ}) = 1$.The restoring torque provided by the torsion band is $	au = C \phi$,where $C = 10^{-6} \, N \cdot m/rad$ and $\phi = 10^{\circ} = 10 	imes \frac{\pi}{180} \, rad$.Equating the torques: $C \phi = N I A B$.Given values: $N = 175$,$I = 1 \, mA = 10^{-3} \, A$,$A = 1 \, cm^2 = 10^{-4} \, m^2$,$C = 10^{-6} \, N \cdot m/rad$,and $\phi = \frac{\pi}{18} \, rad$.Substituting the values: $10^{-6} 	imes \frac{\pi}{18} = 175 	imes 10^{-3} 	imes 10^{-4} 	imes B$.$B = \frac{10^{-6} 	imes \pi}{18 	imes 175 	imes 10^{-7}} = \frac{10 	imes \pi}{18 	imes 175} \approx \frac{31.4}{3150} \approx 0.0099 \approx 10^{-2} \, T$.

Similar Questions

In order to pass $10\,\%$ of the main current through a moving coil galvanometer of $99\,\Omega$,the resistance of the required shunt is ............ $\Omega$.

$A$ galvanometer has a coil of resistance $100 \Omega$ showing a full scale deflection at $50 \mu A$. The resistance that should be added to use it as an ammeter of range $10 \text{ mA}$ is

The current sensitivity of a moving coil galvanometer increases by $20 \%$ when its resistance is doubled. Calculate,by what factor does the voltage sensitivity change?

Two identical galvanometers are converted into a voltmeter and a millivoltmeter. As compared to the series resistance of the voltmeter,the series resistance of the millivoltmeter will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module