At a certain place, the angle of dip is 60^{c | vedclass

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

Question

(D) Given:Horizontal component of earth's magnetic field, $B_H = 0.8 	imes 10^{-4} \,T$Angle of dip, $	heta = 60^{\circ}$Let the earth's total magne

Vedclass · Accepted Answer

$1.6 	imes 10^{-4} \,T$

Vedclass · Answer

(D) Given:Horizontal component of earth's magnetic field, $B_H = 0.8 	imes 10^{-4} \,T$Angle of dip, $	heta = 60^{\circ}$Let the earth's total magnetic field be $B_e$.We know that the horizontal component of the earth's magnetic field is related to the total magnetic field by the formula:$B_H = B_e \cos 	heta$Substituting the given values:$0.8 	imes 10^{-4} = B_e \cos 60^{\circ}$Since $\cos 60^{\circ} = 0.5$ or $\frac{1}{2}$, we have:$0.8 	imes 10^{-4} = B_e 	imes 0.5$Solving for $B_e$:$B_e = \frac{0.8 	imes 10^{-4}}{0.5}$$B_e = 1.6 	imes 10^{-4} \,T$Thus, the earth's overall magnetic field is $1.6 	imes 10^{-4} \,T$.

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

Similar Questions

The magnetic compass is not useful for navigation near the magnetic poles because

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

Give information about Earth's magnetism.

The relation between $Z_E, H_E$,and $B_E$ is . . . . . . .

The relations among the three elements of Earth's magnetic field,namely horizontal component $H$,vertical component $V$,and angle of dip $\delta$ are,($B_{E} =$ total magnetic field):

Vedclass Test Series

Exam Paper Generator

Online Exam Module