Question

Vertical component of Earth's magnetic field is 6× 10⁵ and angle of inclination is 60. calculate the horizontal component of Earth's magnetic field.​

Answer 1

Answer:

Vertical component of Earth's magnetic field is 6× 10⁵ and angle of inclination is 60.

The horizontal component of Earth's magnetic field

Answer 2

Answer:

The horizontal component of the earth's magnetic field's, $B_{H}$, measured is $3.46\times 10^{5} T$.

Explanation:

Given,

The vertical component of the earth's magnetic field, $B_{V}$ = $6\times 10^{5}T$

The angle of inclination, $\theta$ = $60\textdegree$

The earth's magnetic field's horizontal component, $B_{H}$ =?

Let the earth's magnetic field = B

As we know,

• $B_{V} = Bsin\theta$
• B = $\frac{B_{V}}{sin60\textdegree}$              ( $sin60\textdegree = \frac{\sqrt{3} }{2}$ )
• B = $\frac{6\times 10^{5} }{\frac{\sqrt{3} }{2} }$
• B = $\frac{12\times 10^{5} }{1.73}$             ( $\sqrt{3} = 1.73$ )
• B = $6.93\times 10^{5}T$

Therefore, the magnetic field of earth, B = $12\times 10^{5} T$.

As we know,

• $B_{H} = Bcos\theta$
• $B_{H} = 6.93\times 10^{5} cos60\textdegree$
• $B_{H} = 6.93\times 10^{5}\times \frac{1 }{2}$            ( $cos60\textdegree = \frac{1 }{2}$ )
• $B_{H} =$ $3.46\times 10^{5} T$

Hence, the horizontal component of the earth's magnetic field, $B_{H}$ = $3.46\times 10^{5} T$.