Question

A bar magnet of magnetic moment $1.44A{m^2}$ is placed horizontally with the north pole pointing towards north. If the horizontal component of the earth's magnetic field is $18\mu T$ , then the neutral point is a distance (from the magnet) equal to
(A) 10 cm south as well as north
(B) 20 cm east as well as west
(C) $10{\left( 2 \right)^{\dfrac{1}{3}}}$ cm east as well as west
(D) 20 cm south as well as north

Answer 1

Hint : The neutral point lies in the plane bisecting the dipole. By computing the value of the magnetic field and then relating it to the magnetic moment we can find the value of the neutral point from the magnet.

Formula Used: The formulae used in the solution are given here.
$\Rightarrow \overrightarrow B = \dfrac{{{\mu _0}M}}{{4\pi {d^3}}}$ where ${\mu _0}$ is permeability of free space $\left( {4\pi \times {{10}^{ - 7}}} \right)Tm{A^{ - 1}}$ , $B$ is the Magnetic Field, $M$ is the magnetic moment and $d$ is the distance of separation.

Complete step by step answer
As we all know, a magnet has two poles, i.e., North and South.
A magnetic field is one that describes the magnetic influence on moving electric charges. It can be defined as a vector field in the vicinity of a magnet, electric current, or changing electric field, in which a charged particle is under the influence of magnetic forces. Magnetic fields force moving electrically charged particles in a circular or helical path.
Magnetic fields such as that of Earth cause magnetic compass needles and other permanent magnets to line up in the direction of the field.
Magnetic moment can be defined as the magnetic strength and orientation of a magnet or other object that produces a magnetic field.
It is known that, for a magnetic dipole with its pole pointing to the north, the neutral point in the broadsides in position.
We know that, $\overrightarrow B = \dfrac{{{\mu _0}M}}{{4\pi {d^3}}}$ where ${\mu _0}$ is permeability of free space $\left( {4\pi \times {{10}^{ - 7}}} \right)Tm{A^{ - 1}}$ , $B$ is the Magnetic Field, $M$ is the magnetic moment and $d$ is the distance of separation.
It is given that, $\overrightarrow B = 18\mu T$ and ${\mu _0}M = 1.44 \times {10^{ - 7}}A{m^2}$
Substituting these values in the equation,
$\Rightarrow \dfrac{{1.44 \times {{10}^{ - 7}}}}{{4\pi {d^3}}} = 18 \times {10^{ - 6}}$
To find the neutral position, we have to find the value of $d$ .
Simplifying the equation we get,
$\Rightarrow \dfrac{{1.44 \times {{10}^{ - 7}}}}{{18 \times {{10}^{ - 6}} \times 4\pi }} = {d^3}$
On taking cube roots on both sides we get,
$\Rightarrow {\left[ {\dfrac{{1.44 \times {{10}^{ - 7}}}}{{18 \times {{10}^{ - 6}} \times 4\pi }}} \right]^{\dfrac{1}{3}}} = d$
Thus, calculating the value of $d$ , we find
$\Rightarrow {d^3} = 8 \times {10^{ - 3}}$
$\Rightarrow d = 2 \times {10^{ - 1}}m$
Thus, $d = 20cm$ . It has been said earlier that, neutral point is lying on the broadside.
$\therefore$ The neutral point is 20 cm east as well as west.
The correct option is Option B.

Note
The characteristics of the current loop are summarized in its magnetic moment. Magnetic fields are produced by electric fields, which can be current in wires, or current associated with electrons in their orbits.