$A$ 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 $B_H = 3.6 \times 10^{-5} \, T$)

  • A
    $14.6$
  • B
    $19.4$
  • C
    $9.7$
  • D
    $4.9$

Explore More

Similar Questions

The error in measuring the current with a tangent galvanometer is minimum when the deflection is about $45^o$. (in $^o$)

Difficult
View Solution

$A$ bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

The unit of the reduction factor of a tangent galvanometer is

In the sum and difference method of a vibration magnetometer,the time period is greater if:

The ratio of magnetic moments of two bar magnets is $13 : 5$. These magnets are held together in a vibration magnetometer and are allowed to oscillate in the Earth's magnetic field. When like poles are together,$15$ oscillations per minute are made. What will be the frequency of oscillation of the system if unlike poles are together? (in $oscillations/min$)

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