A circular coil of 50 turns and radius 10 cm carrying a current of 5.0 A is suspendedvertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field linesmake an angle of 90° with the normal of the coil. Calculate the magnitude of thecounter torque that must be applied to prevent the coil from turning?

Correct option is B

$$\textbf{Given:} \\\text{Number of turns in the coil, } N = 50 \\\text{Radius of the coil, } r = 10\, \text{cm} = 0.1\, \text{m} \\\text{Current through the coil, } I = 5\, \text{A} \\\text{Magnetic field strength, } B = 1\, \text{T} \\\theta = 90^\circ \\\text{The torque experienced by a current-carrying coil in a magnetic field is given by:} \\\tau = N \cdot I \cdot A \cdot B \cdot \sin\theta \\\text{Where } A = \pi r^2 \\\text{Since } \theta = 90^\circ, \sin 90^\circ = 1 \\\tau = 50 \cdot 5 \cdot (\pi \cdot 0.1^2) \cdot 1 \cdot \sin 90^\circ \\\tau = 50 \cdot 5 \cdot \pi \cdot 0.01 \\\tau = 7.85\, \text{Nm} \\\boxed{\text{Counter torque} = 7.85\, \text{Nm}}$$​​