Question

A straight conductor carrying a current is kept in a uniform magnetic field so as to experience maximum force. If now the conductor is turned in its own plane such that the force acting on it is half of the maximum force, then the angle made by the conductor in the final position with respect to the field is:

• A. ${60}^0$
• B. ${45}^0$
• C. ${30}^0$
• D. ${90}^0$

Answer 1

Answer: (C)

${30}^0$

Max force is given by: $F=Bil$
At half the force, $F^{\prime }=Bilsin\theta =\frac{F}{2}$
$Bilsin\theta =\frac{Bil}{2}$
$\theta ={30}^{\circ }$