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

If $B_V$ and $B_H$ are respectively the vertical and horizontal components of the earth's magnetic field at a place where the angle of dip is $60^{\circ}$, then the total magnetic field at that place is

The plane of a dip circle is set in the geographic meridian and the apparent dip is $\delta_1$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $\delta_2$. The declination $\theta$ at the place is

$A$ dip circle is at a right angle to the magnetic meridian. What will be the apparent dip in degrees?

At a point $A$ on the earth's surface, the angle of dip is $\delta = +25^{\circ}$. At a point $B$ on the earth's surface, the angle of dip is $\delta = -25^{\circ}$. We can interpret that:

Define the angle of dip and the angle of declination.

Vedclass Test Series

Exam Paper Generator

Online Exam Module