$A$ tangent galvanometer has a coil of $25$ turns and a radius of $15\, cm$. The horizontal component of the earth's magnetic field is $3 \times 10^{-5}\, T$. The current required to produce a deflection of $45^{\circ}$ in it is....$A$

  • A
    $0.29$
  • B
    $1.2$
  • C
    $3.6 \times 10^{-5}$
  • D
    $0.14$

Explore More

Similar Questions

The radius of the coil of a Tangent galvanometer,which has $10$ turns,is $0.1\,m$. The current required to produce a deflection of $60^{\circ}$ $(B_H = 4 \times 10^{-5}\,T)$ is.....$A$

$A$ short bar magnet is placed in the magnetic meridian of the earth with its north pole pointing towards the geographic north. Neutral points are found at a distance of $30 \, cm$ from the magnet on the East-West line, drawn through the center of the magnet. The magnetic moment of the magnet in $A \cdot m^2$ is close to: (Given $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$ and $B_H = 3.6 \times 10^{-5} \, T$)

When the $N$-pole of a bar magnet points towards the geographic south and the $S$-pole points towards the geographic north,the null points are located at the:

If a brass bar is placed on a vibrating magnet, then its time period

The moment of inertia of a magnetic needle is $40 \, g \cdot cm^2$ and it has a time period of $3 \, s$ in the Earth's horizontal magnetic field of $3.6 \times 10^{-5} \, Wb/m^2$. Its magnetic moment will be (in $A \cdot m^2$):

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo