Question

A coil of N turns and radius r carrying a current i is rotated in a
uniform magnetic field of strength B. In a particular position of the
coil, the torque acting on it is half of the maximum torque that
can act on it. What is the angle between the magnetic field and
the plane of the coil ?:
(a) 600
(b) 300
(C) 450
(d) 90°​

Answer 1

Answer:

300

hope it helps you........

pls mark as brainelist

Answer 2

Given:

A coil of N turns and radius r carrying a current i is rotated in a uniform magnetic field of strength B. In a particular position of the coil, the torque acting on it is half of the maximum torque that

can act on it.

To find:

Angle between Magnetic field and plane of coil.

Calculation:

Let angle between Magnetic Moment and Magnetic Field Intensity be $\theta$.

$\rm{ \tau = MB \times \sin( \theta) }$

The torque experience at this position is half as that of the maximum torque.

$\rm{ \tau = \dfrac{1}{2} MB }$

$\rm{ = > MB \times \sin( \theta) = \dfrac{1}{2} MB }$

$\rm{ = > \sin( \theta) = \dfrac{1}{2} }$

$= > \: \theta = 30 \degree$

Hence angle between plane of the coil and magnetic field intensity is

$\boxed{ \phi = 90\degree - \theta= 60 \degree}$