The value of horizontal component of earth's | vedclass

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

Question

(D) The horizontal component of the earth's magnetic field is given by $B_H = 0.35 	imes 10^{-4} \,T$.The angle of dip is $\delta = 60^{\circ}$.The r

Vedclass · Accepted Answer

$0.61$

Vedclass · Answer

(D) The horizontal component of the earth's magnetic field is given by $B_H = 0.35 	imes 10^{-4} \,T$.The angle of dip is $\delta = 60^{\circ}$.The relationship between the vertical component $(B_V)$,horizontal component $(B_H)$,and the angle of dip $(\delta)$ is given by the formula: $B_V = B_H 	an \delta$.Substituting the given values into the formula:$B_V = (0.35 	imes 10^{-4}) 	imes 	an(60^{\circ})$.Since $	an(60^{\circ}) = \sqrt{3} \approx 1.732$,we have:$B_V = 0.35 	imes 1.732 	imes 10^{-4} \,T$.$B_V \approx 0.6062 	imes 10^{-4} \,T$.Rounding to two decimal places,we get $B_V \approx 0.61 	imes 10^{-4} \,T$.Therefore,the correct option is $D$.

The value of horizontal component of earth's magnetic field at a place is $0.35 \times 10^{-4} \,T$. If the angle of dip is $60^{\circ}$,the value of vertical component of earth's magnetic field is nearly ............. $\times 10^{-4} \,T$.

Similar Questions

The angle of dip is the angle

$A$ compass needle will show which one of the following directions at the Earth's magnetic pole?

The lines joining the places of the same horizontal intensity are known as

At which place does the Earth's magnetism become horizontal?

What is magnetic declination?

Vedclass Test Series

Exam Paper Generator

Online Exam Module