Question

calculate the earths magnetic field at a place where angle of dip 50 degree and horizontal component of earths magnetic field is 0.3G
​

Answer 1

Answer:

$\boxed{\mathfrak{Earth's \ magnetic \ field \ (B_e) = 0.5 \ G}}$

Given:

Angle of dip ($\rm \delta$) = 50°

Horizontal component of earth's magnetic field ($\rm B_H$) = 0.3 G

Explanation:

Relation between earth's magnetic field and horizontal component of earth's magnetic field is given as:

$\boxed{ \bold{B_H = B_e \: cos \delta}}$

By substituting values we get:

$\rm \implies 0.3 = B_e \: cos 50 \degree \\ \\ \rm \implies 0.3 = B_e \times 0.6 \\ \\ \rm \implies B_e = \frac{0.3}{0.6} \\ \\ \rm \implies B_e = 0.5 \: G$

Answer 2

Answer:

Given that,

Total strength of field B=0.5G,

Horizontal component of the field H=0.3G,

Total strength of field is given by

B=

(H

2

+V

2

)

​

where, V is the vertical component of the field.

B

2

=H

2

+V

2

V

2

=B

2

−H

2

=(0.5)

2

−(0.3)

2

=0.4G

Now, angle of dip is

tanδ=

H

V

​

=

0.3

0.4

​

=

3

4

​

δ=tan

−1

3

4

​