Question

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14\text{cm}$ from the centre of the magnet. The earth's magnetic field at the place is $0.36\text{G}$ and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point $\left(\text{i.e.,}14\text{cm}\right)$ from the center of the magnet?

• A. $0.36\text{G}$
• B. $0.44\text{G}$
• C. $0.72\text{G}$
• D. $0.54\text{G}$

Answer 1

The correct option is D $0.54\text{G}$

Since, dip angle $\delta =0^{\circ }$ thus, the horizontal component of the Earth's magnetic field at the given place will be equal to the total Earth's magnetic field,i.e,

$B_H=0.36\text{G}$.

At the null point (along the axis), earth's magnetic field and bar's magnetic field are opposite in direction.

$B_{axial}=B_H.........\left(1\right)$

Where,
$B_{axial}=$ external magnetic field due to bar magnet
$B_H=$ horizontal component of earth's magnetic field

Thus, the magnetic field at the axial point of the bar magnet is given by,

$B_{axial}=\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}..........\left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$,

$B_{axial}=\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}=B_H$

The magnetic field due to a short bar magnet at any point on the axial line is twice the magnetic field at a point on the equatorial line of that magnet at the same distance. So,

$B_{equatorial}=\frac{{\mu }_{0}M}{4\pi d^3}=\frac{B_H}{2}$


At the equatorial line, the directions of magnetic field due to bar magnet and Earth's magnetic field are the same.

So, the total magnetic field is,

$B=B_H+B_{equatorial}=B_H+\frac{B_H}{2}$

$\therefore B=0.36+0.18=0.54\text{G}$

Hence, the magnetic field is $0.54\text{G}$ in the direction of earth's magnetic field.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.

Why this question: Null points: At null points, the field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field. (i.e.,) $B_{magnet}=B_H$ Where, $B_{magnet}=\text{external magnetic field due to magnet and}{B}_H=\text{horizontal component of earth's magnetic field}.$