Question

A magnetic dipole of magnetic moment 0.72√2 A m2 is placed horizontally with the north pole pointing towards east. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 μT.

Answer 1

Explanation:

To find: The position of the neutral point if the horizontal component of the earth's magnetic field is 18 μT.

Step 1:

Given data in the question:

Magnetic moment of a dipolar magnet,  $M=0.72 \sqrt{2} \mathrm{A}-\mathrm{m}^{2}$

Horizontal component of the magnetic field on earth, BH = 18 μT

Let d be the neutral point distance from a dipole.

Step 2:

The equatorial dipole line is given by,

$B=\frac{\mu_{0} M}{4 \pi d^{3}}$

Step 3:

On substituting the values, we get

$\frac{10^{-7} \times 4 \pi}{4 \pi} \times \frac{0.72 \sqrt{2}}{d^{3}}=18 \times 10^{-6}$

    $10^{-7} \times \frac{0.72 \sqrt{2}}{d^{3}}=18 \times 10^{-6}$

$10^{-7} \times \frac{0.72 \times 1.414}{d^{3}}=18 \times 10^{-6}$

                       $d^{3}=\frac{10^{-7} \times 1.414 \times 0.72}{18 \times 10^{-6}}$

                      $d^{3}=\frac{10^{-7} \times 1.018}{18 \times 10^{-6}}$

                      $d^{3}=\frac{10^{-1} \times 1.018}{18}$

                      $d^{3}=10^{-1} \times 0.05656$

                      $d^{3}=0.005656$

                     $d=\sqrt[8]{0.005656}$

                    $d \approx 0.2 m=20 \mathrm{cm}$