A circular coil of 50 turns and a radius of 10 cm carrying a current of 5.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 90° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning?
Coil Torque Calculation: Understanding Magnetic Field Effects
This problem involves calculating the torque experienced by a current-carrying circular coil when placed in a uniform magnetic field. The coil tends to align its magnetic moment with the external magnetic field. To prevent this turning, an equal and opposite (counter) torque must be applied.
Circular Coil Parameters and Given Values
Let's list the important parameters provided for the circular coil:
- Number of turns (\(N\)): \(50\)
- Radius (\(r\)): \(10 \text{ cm} = 0.1 \text{ m}\)
- Current (\(I\)): \(5.0 \text{ A}\)
- Magnetic field strength (\(B\)): \(1.0 \text{ T}\)
- Angle between the magnetic field lines and the normal of the coil (\(\theta\)): \(90^\circ\)
Magnetic Moment of the Coil
First, we need to calculate the area (\(A\)) of the circular coil. For a circle, the area is given by the formula:
\[A = \pi r^2\]
Substitute the given radius:
\[A = \pi (0.1 \text{ m})^2\]
\[A = 0.01 \pi \text{ m}^2\]
Next, we calculate the magnetic dipole moment (\(M\)) of the circular coil. The magnetic moment for a coil with \(N\) turns, carrying current \(I\), and having area \(A\) is given by:
\[M = NIA\]
Substitute the values of \(N\), \(I\), and \(A\):
\[M = 50 \times 5.0 \text{ A} \times (0.01 \pi \text{ m}^2)\]
\[M = 250 \times 0.01 \pi \text{ A} \cdot \text{m}^2\]
\[M = 2.5 \pi \text{ A} \cdot \text{m}^2\]
Using the approximate value of \(\pi \approx 3.14159\):
\[M \approx 2.5 \times 3.14159 \text{ A} \cdot \text{m}^2\]
\[M \approx 7.853975 \text{ A} \cdot \text{m}^2\]
Torque Calculation and Counter Torque
The torque (\(\tau\)) experienced by a magnetic dipole (like our circular coil) in a uniform magnetic field is given by the formula:
\[\tau = MB \sin\theta\]
Here, \(\theta\) is the angle between the magnetic moment vector (which is normal to the coil's plane) and the magnetic field vector. In this problem, it is stated that the field lines make an angle of \(90^\circ\) with the normal of the coil. Therefore, \(\theta = 90^\circ\).
Substitute the values of \(M\), \(B\), and \(\theta\):
\[\tau = (2.5 \pi \text{ A} \cdot \text{m}^2) \times (1.0 \text{ T}) \times \sin(90^\circ)\]
Since \(\sin(90^\circ) = 1\):
\[\tau = 2.5 \pi \times 1.0 \times 1 \text{ Nm}\]
\[\tau = 2.5 \pi \text{ Nm}\]
Calculating the numerical value:
\[\tau \approx 2.5 \times 3.14159 \text{ Nm}\]
\[\tau \approx 7.853975 \text{ Nm}\]
To prevent the coil from turning, a counter torque of the same magnitude must be applied. Therefore, the magnitude of the counter torque is approximately \(7.85 \text{ Nm}\).
| Parameter | Symbol | Value | Unit |
|---|---|---|---|
| Number of turns | \(N\) | \(50\) | - |
| Radius | \(r\) | \(0.1\) | m |
| Current | \(I\) | \(5.0\) | A |
| Magnetic Field | \(B\) | \(1.0\) | T |
| Angle | \(\theta\) | \(90\) | degrees |
| Area | \(A\) | \(0.01\pi\) | m\(^2\) |
| Magnetic Moment | \(M\) | \(2.5\pi\) | A·m\(^2\) |
| Torque | \(\tau\) | \(2.5\pi \approx 7.85\) | Nm |
The calculated magnitude of the counter torque is approximately \(7.85 \text{ Nm}\).
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