Question:
A short bar magnet placed in a horizontal plane has its axis aligned along north-south direction. Null points are found on the axis of the magnet at \(20\ \text{cm}\) from the center of magnet. The earth's magnetic field at the place is \(B\) and angle of dip is \(0^\circ\). If the total magnetic field on the normal bisector of the magnet \(20\ \text{cm}\) from the center of the magnet is \(0.6\ \text{G}\), then the magnitude of \(B\) is:
A short bar magnet placed in a horizontal plane has its axis aligned along north-south direction. Null points are found on the axis of the magnet at \(20\ \text{cm}\) from the center of magnet. The earth's magnetic field at the place is \(B\) and angle of dip is \(0^\circ\). If the total magnetic field on the normal bisector of the magnet \(20\ \text{cm}\) from the center of the magnet is \(0.6\ \text{G}\), then the magnitude of \(B\) is:
Show Hint
For a short bar magnet at the same distance,
\[
B_{\text{axial}}=2B_{\text{equatorial}}
\]
This relation is very useful in problems involving null points and magnetic field on the normal bisector.
Updated On: Jun 26, 2026
- \(0.2\ \text{G}\)
- \(0.4\ \text{G}\)
- \(1.2\ \text{G}\)
- \(0.3\ \text{G}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the given condition for null points.
The bar magnet is placed along the north-south direction.
The angle of dip is \[ 0^\circ \] So the earth's magnetic field is horizontal and is denoted by \[ B \] Null points are found on the axial line of the magnet at a distance \[ r=20\ \text{cm} \] At a null point, the magnetic field due to the magnet is equal in magnitude and opposite in direction to the earth's magnetic field.
Therefore, on the axial line, \[ B_{\text{axial}}=B \]
Step 2: Use the relation between axial and equatorial magnetic fields.
For a short bar magnet, the magnetic field on the axial line is \[ B_{\text{axial}}=\frac{\mu_0}{4\pi}\frac{2M}{r^3} \] The magnetic field on the equatorial line or normal bisector is \[ B_{\text{equatorial}}=\frac{\mu_0}{4\pi}\frac{M}{r^3} \] Hence, \[ B_{\text{axial}}=2B_{\text{equatorial}} \] Therefore, \[ B_{\text{equatorial}}=\frac{B_{\text{axial}}}{2} \] Since \[ B_{\text{axial}}=B \] we get \[ B_{\text{equatorial}}=\frac{B}{2} \]
Step 3: Find the total magnetic field on the normal bisector.
On the normal bisector, the field due to the magnet is opposite to the magnetic moment direction.
The earth's magnetic field and the magnetic field due to the magnet act in the same line, so the resultant field is \[ B_{\text{total}}=B+B_{\text{equatorial}} \] Substituting \[ B_{\text{equatorial}}=\frac{B}{2} \] we get \[ B_{\text{total}}=B+\frac{B}{2} \] \[ B_{\text{total}}=\frac{3B}{2} \] Given, \[ B_{\text{total}}=0.6\ \text{G} \] Thus, \[ \frac{3B}{2}=0.6 \]
Step 4: Calculate the value of \(B\).
\[ B=0.6\times\frac{2}{3} \] \[ B=0.4\ \text{G} \]
Step 5: Final conclusion.
Therefore, the magnitude of earth's magnetic field is \[ \boxed{0.4\ \text{G}} \]
The bar magnet is placed along the north-south direction.
The angle of dip is \[ 0^\circ \] So the earth's magnetic field is horizontal and is denoted by \[ B \] Null points are found on the axial line of the magnet at a distance \[ r=20\ \text{cm} \] At a null point, the magnetic field due to the magnet is equal in magnitude and opposite in direction to the earth's magnetic field.
Therefore, on the axial line, \[ B_{\text{axial}}=B \]
Step 2: Use the relation between axial and equatorial magnetic fields.
For a short bar magnet, the magnetic field on the axial line is \[ B_{\text{axial}}=\frac{\mu_0}{4\pi}\frac{2M}{r^3} \] The magnetic field on the equatorial line or normal bisector is \[ B_{\text{equatorial}}=\frac{\mu_0}{4\pi}\frac{M}{r^3} \] Hence, \[ B_{\text{axial}}=2B_{\text{equatorial}} \] Therefore, \[ B_{\text{equatorial}}=\frac{B_{\text{axial}}}{2} \] Since \[ B_{\text{axial}}=B \] we get \[ B_{\text{equatorial}}=\frac{B}{2} \]
Step 3: Find the total magnetic field on the normal bisector.
On the normal bisector, the field due to the magnet is opposite to the magnetic moment direction.
The earth's magnetic field and the magnetic field due to the magnet act in the same line, so the resultant field is \[ B_{\text{total}}=B+B_{\text{equatorial}} \] Substituting \[ B_{\text{equatorial}}=\frac{B}{2} \] we get \[ B_{\text{total}}=B+\frac{B}{2} \] \[ B_{\text{total}}=\frac{3B}{2} \] Given, \[ B_{\text{total}}=0.6\ \text{G} \] Thus, \[ \frac{3B}{2}=0.6 \]
Step 4: Calculate the value of \(B\).
\[ B=0.6\times\frac{2}{3} \] \[ B=0.4\ \text{G} \]
Step 5: Final conclusion.
Therefore, the magnitude of earth's magnetic field is \[ \boxed{0.4\ \text{G}} \]
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