Question

Two identical short bar magnets, each having a magnetic moment of \(1 \, {A} \cdot {m}^2\), are placed 2 m apart between their centers with their axes perpendicular to each other. The net magnetic field at the midpoint of the line joining the centers of the two magnets is:

• A. \(2 \times 10^{-7} \, {T}\)
• B. \(1 \times 10^{-7} \, {T}\)
• C. \(\sqrt{5} \times 10^{-7} \, {T}\)
• D. \(\sqrt{3} \times 10^{-7} \, {T}\)

Hint:

Magnetic field due to a bar magnet varies along axial and equatorial lines; vector sum applies for perpendicular fields.

Answer 1

The Correct Option is C

The magnetic field \( B \) due to a short bar magnet at a point along its axial line at a distance \( r \) is:

\[ B = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \]
where
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot \text{m/A} \) (permeability of free space),
- \( M = 1 \, \text{A} \cdot \text{m}^2 \) (magnetic moment),
- \( r = 1 \, \text{m} \) (distance from the center to midpoint, since total distance is 2 m).

Calculate the magnetic field from each magnet at midpoint:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \frac{2 \times 1}{1^3} = 2 \times 10^{-7} \, \text{T} \]
Since the magnets are perpendicular, their fields at the midpoint are perpendicular vectors, each of magnitude \( 2 \times 10^{-7} \, \text{T} \).

The resultant magnetic field \( B_{\text{net}} \) is the vector sum:
\[ B_{\text{net}} = \sqrt{(2 \times 10^{-7})^2 + (2 \times 10^{-7})^2} = \sqrt{2 \times (2 \times 10^{-7})^2} = 2 \times 10^{-7} \sqrt{2} \]

However, the given answer is \( \sqrt{5} \times 10^{-7} \, \text{T} \), so let's verify if one magnet contributes axial field and the other contributes equatorial field.

Magnetic field on the axial line:
\[ B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} = 2 \times 10^{-7} \, \text{T} \]
Magnetic field on the equatorial line:
\[ B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{r^3} = 1 \times 10^{-7} \, \text{T} \]

Since the magnets are perpendicular, one magnet's field at midpoint is axial, the other's equatorial. Thus,
\[ B_{\text{net}} = \sqrt{(2 \times 10^{-7})^2 + (1 \times 10^{-7})^2} = \sqrt{4 + 1} \times 10^{-14} = \sqrt{5} \times 10^{-7} \, \text{T} \]