The angle of dip at a certain place on earth is $60^{\circ}$ and the magnitude of earth's horizontal component of magnetic field is $0.26 \, G$. The magnetic field at the place on earth is.....$G$

  • A
    $0.13$
  • B
    $0.26$
  • C
    $0.52$
  • D
    $0.65$

Explore More

Similar Questions

At the magnetic poles of the Earth,the angle of dip is .....$^o$.

Which of the following statements proves that Earth has a magnetic field?

The angles of dip are $30^{\circ}$ and $45^{\circ}$ at two different places. The ratio of the horizontal components of the Earth's magnetic field at these places will be:

The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} \; Wb/m^2$. If the angle of dip at that place is $60^{\circ}$,then the value of the vertical component of the earth's magnetic field will be ........ $\times 10^{-4} \; Wb/m^2$.

If $\theta_1$ and $\theta_2$ are the apparent angles of dip observed in two vertical planes at right angles to each other,then the true angle of dip $\theta$ is given by:

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