Torque on a Current-Carrying Rectangular Loop in a Magnetic Field
Consider a rectangular loop of sides \( a \) and \( b \) carrying a current \( I \), placed in a uniform magnetic field \( \vec{B} \). Let the area vector of the loop be \( \vec{A} \), making an angle \( \theta \) with \( \vec{B} \).
- The magnetic forces on opposite sides of the loop produce a couple that tends to rotate the loop.
- The forces on the sides parallel to \( \vec{B} \) are zero because \( \vec{F} = I \vec{L} \times \vec{B} \) and \( \vec{L} \parallel \vec{B} \) gives zero cross product.
- The forces on the sides perpendicular to \( \vec{B} \) produce a torque about the axis through the center of the loop.
Step 1: Magnetic Dipole Moment
The magnetic dipole moment of the loop is defined as:
\[
\vec{m} = I \vec{A}
\]
where \( \vec{A} \) is a vector perpendicular to the plane of the loop, with magnitude equal to the area of the loop: \( A = a b \).
Step 2: Torque on the Loop
The torque \( \vec{\tau} \) experienced by a current loop in a magnetic field is given by:
\[
\vec{\tau} = \vec{m} \times \vec{B}
\]
The magnitude of the torque is:
\[
\tau = m B \sin \theta = I A B \sin \theta
\]
where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).
- This torque tends to rotate the loop so that the area vector aligns with the magnetic field.
- The direction of torque follows the right-hand rule, perpendicular to both \( \vec{m} \) and \( \vec{B} \).
Step 3: Summary
- Magnetic dipole moment: \( \vec{m} = I \vec{A} \)
- Torque on the loop: \( \vec{\tau} = \vec{m} \times \vec{B} \)
- Magnitude: \( \tau = I A B \sin \theta \)
This shows that the torque depends on the current, area of the loop, magnetic field strength, and orientation of the loop with respect to the field.