Question

At a place, the earth's magnetic field is B and angle of dip is 60°. What is the value of horizontal component of earth magnetic field at that place

Answer 1

The correct answer to the question is $\frac{B}{2}$

CALCULATION:

As per the question, the magnetic field of earth is given as B.

The angle of dip is given as $\theta$ = 60 degree.

We are asked to calculate the horizontal component of the earth magnetic field.

The angle of dip is the angle made by the horizontal with the earth magnetic field.

Hence, the horizontal component of the earth magnetic field is calculated as -

Horizontal component $B_{H} =\ Bcos\theta$

                                                        = $B\times cos60$

                                                        = $B\times \frac{1}{2}$

                                                        = $\frac{B}{2}$        [ans]

Hence, the horizontal component of the earth magnetic field at the place is $\frac{B}{2}$ .

Answer 2

Given:

Magnetic field B

Angle of dip 60 degrees

To find:

The value of horizontal component of earth magnetic field at that place

Solution:

Angle of dip: The angle made by the horizontal axis of the component with respect to the earth's magnetic field is referred to as the angle of dip.

To find the angle of the horizontal component,

Bh = B cos θ

Substituting the value for Θ,

Bh = B cos 60

cos 60 = 1/2

Bh = B * 1/2

Therefore,

We get,

B/2

Hence, the value of horizontal component of earth magnetic field at that place is B/2.