Question

Two short magnetic dipoles \(m_1\) and \(m_2\) each having magnetic moment of \(1 \text{ Am}^2\) are placed at point O and P respectively. The distance between OP is 1 meter. The torque experienced by the magnetic dipole \(m_2\) due to the presence of \(m_1\) is _________ \(\times 10^{-7} \text{ Nm}\).
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Hint:

Equatorial field is half the axial field for the same distance. Always determine the field direction first to find the torque angle correctly.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
A magnetic dipole \(m_1\) creates a magnetic field \(B_1\) in its surrounding space. Another dipole \(m_2\) placed in this field experiences a torque given by \(\vec{\tau} = \vec{m_2} \times \vec{B_1}\).
Step 2: Key Formula or Approach:
1. Magnetic field of a short dipole at an equatorial point: \(B = \frac{\mu_0}{4\pi} \frac{m}{r^3}\), directed opposite to \(\vec{m}\).
2. Torque: \(\tau = m_2 B_1 \sin \theta\).
Step 3: Detailed Explanation:
Given: \(m_1 = m_2 = 1 \text{ Am}^2\), \(r = 1 \text{ m}\).
From the figure, point P is on the equatorial line of dipole \(m_1\).

1. Magnetic field \(\vec{B_1}\) at point P due to \(m_1\):
Since P is on the equatorial plane, the magnitude is:
\[ B_1 = \frac{\mu_0}{4\pi} \frac{m_1}{r^3} = 10^{-7} \times \frac{1}{1^3} = 10^{-7} \text{ T} \]
The direction of \(\vec{B_1}\) is opposite to \(\vec{m_1}\) (pointing downwards if \(\vec{m_1}\) is upwards).

2. Torque on \(m_2\):
Dipole \(m_2\) is oriented horizontally (to the right). The magnetic field \(\vec{B_1}\) is vertical.
Thus, the angle \(\theta\) between \(\vec{m_2}\) and \(\vec{B_1}\) is \(90^\circ\).
\[ \tau = m_2 B_1 \sin 90^\circ = 1 \times 10^{-7} \times 1 = 10^{-7} \text{ Nm} \]

Comparing with the form \(\text{value} \times 10^{-7}\), the value is 1.
Step 4: Final Answer:
The torque is \(1 \times 10^{-7}\) Nm.