CBSE · Class 12 · Physics · Moving Charges and MagnetismA circular coil of 30 turns and radius 8.0 cm is carrying a current of 6.0 A and is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of $60^\circ$ with the normal of the coil. Calculate the magnitude of the counter-torque that must be applied to prevent the coil from turning.
To calculate the counter-torque required to prevent the coil from turning, we need to determine the magnetic torque acting on the current-carrying coil placed in a magnetic field.
Given Data:
- Number of turns ($N$) = $30$
- Radius of the coil ($R$) = $8.0 \text{ cm} = 0.08 \text{ m}$
- Current in the coil ($I$) = $6.0 \text{ A}$
- Magnetic field magnitude ($B$) = $1.0 \text{ T}$
- Angle between the magnetic field and the normal of the coil ($\theta$) = $60^\circ$
Step 1: Calculate the area of the circular coil ($A$) $$A = \pi R^2$$ $$A = 3.1416 \times (0.08 \text{ m})^2$$ $$A = 3.1416 \times 0.0064 \text{ m}^2 = 0.020106 \text{ m}^2$$
Step 2: Calculate the magnetic dipole moment of the coil ($M$)\nThe magnetic moment is given by the formula: $$M = N \cdot I \cdot A$$\nSubstitute the known values: $$M = 30 \times 6.0 \text{ A} \times 0.020106 \text{ m}^2$$ $$M = 180 \times 0.020106 = 3.619 \text{ A}\cdot\text{m}^2$$
Step 3: Calculate the magnetic torque ($\tau$) acting on the coil\nThe torque experienced by a magnetic dipole in a uniform magnetic field is given by: $$\tau = M B \sin\theta$$ (Note: $\theta$ is the angle between the magnetic field vector and the area vector/normal, so we use $\sin\theta$ when $\theta$ is given with respect to the normal)\nSubstitute the values into the torque equation: $$\tau = (3.619 \text{ A}\cdot\text{m}^2) \times (1.0 \text{ T}) \times \sin(60^\circ)$$ $$\tau = 3.619 \times \frac{\sqrt{3}}{2}$$ $$\tau = 3.619 \times 0.866 = 3.134 \text{ N}\cdot\text{m}$$
Step 4: Determine the counter-torque\nTo prevent the coil from turning, an equal and opposite counter-torque must be applied. $$\text{Counter-torque} = \tau = 3.13 = 3.13 \text{ N}\cdot\text{m}$$
Answer:\nThe magnitude of the counter-torque that must be applied is $3.13 \text{ N}\cdot\text{m}$.