A dip circle is adjusted so that its needle m | vedclass

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

Question

(C) The true angle of dip $	heta$ is given by $	an 	heta = \frac{B_V}{B_H}$,where $B_V$ is the vertical component and $B_H$ is the horizontal compo

Vedclass · Accepted Answer

More than $40^{\circ}$

Vedclass · Answer

(C) The true angle of dip $	heta$ is given by $	an 	heta = \frac{B_V}{B_H}$,where $B_V$ is the vertical component and $B_H$ is the horizontal component of the Earth's magnetic field.When the dip circle is rotated by an angle $\alpha = 30^{\circ}$ from the magnetic meridian,the horizontal component $B_H$ changes to $B'_H = B_H \cos \alpha$,while the vertical component $B_V$ remains unchanged.The apparent angle of dip $	heta'$ is given by $	an 	heta' = \frac{B_V}{B'_H} = \frac{B_V}{B_H \cos \alpha}$.Substituting $	an 	heta = \frac{B_V}{B_H}$,we get $	an 	heta' = \frac{	an 	heta}{\cos 30^{\circ}}$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}  	an 	heta$.Therefore,$	heta' > 40^{\circ}$. The needle will dip by an angle greater than $40^{\circ}$.

Similar Questions

The values of the apparent angles of dip in two planes at right angles to each other are $45^{\circ}$ and $30^{\circ}$ respectively. The true value of the angle of dip at the place is ............

The angle of dip is $90^{\circ}$ at:

At a certain place, the angle of dip is $60^{\circ}$ and the horizontal component of the earth's magnetic field $(B_H)$ is $0.8 \times 10^{-4} \,T$. The earth's overall magnetic field is

The magnetic field of the Earth is due to

Vedclass Test Series

Exam Paper Generator

Online Exam Module