
A current carrying loop is free to turn in a uniform magnetic field. The loop will then come into equilibrium when its plane is inclined at.
A) ${0^ \circ }$ to the direction of field
B) ${45^ \circ }$ to the direction of field
C) ${90^ \circ }$ to the direction of field
D) ${60^ \circ }$ to the direction of field
Answer
124.5k+ views
Hint:At equilibrium the torque acting on it will be zero. We know that torque is the cross product of the magnetic moment of the current carrying coil and magnetic field. For torque to be zero the angle between the area vector to the plane and the direction of the magnetic field should be zero. Using this we can find the orientation of the plane with respect to the magnetic field.
Complete step by step solution:
It is given that a current carrying loop is free to turn in a uniform magnetic field.
We are asked to find the inclination of the plane when it comes to equilibrium. We know that the condition for equilibrium is that the net force acting on it should be zero
Here since rotation is involved we can say the net torque should be zero at equilibrium.
Torque in the case of a current carrying coil is related to magnetic moment M and magnetic field B as
$\tau = M \times B = MB\sin \theta $
Magnetic moment of current carrying coil is given as
$M = NiA$
Where $N$ is the number of turns,$i$ is current through the coil and $A$ is the area of cross section.
Therefore we can write
$\tau = NiAB\sin \theta $
For torque to be zero the sin$\theta $ should be zero.
$\sin \theta $ will be zero when $\theta = {0^ \circ }$
Here we need to know that $\theta $ is the angle made by the area vector with the magnetic field.
Area vector is the normal vector to the surface of the area. That is, the angle between the perpendicular
to the plane and magnetic field should be zero. Which means the area vector will be along the direction of the field.
Therefore, the angle between plane and magnetic field should be perpendicular to each other so that the area vector will be along the direction of the field.
So, The correct answer is option C.
Note: Remember that the angle $\theta $ that appears in the equation for torque is the angle between area vector and the magnetic field. In the question we are asked to find the angle made by the plane with the magnetic field. Since the area vector is always directed perpendicular to the plane, if the angle made by it is ${0^ \circ }$ then the plane will be making an angle ${90^ \circ }$ to the field.
Complete step by step solution:
It is given that a current carrying loop is free to turn in a uniform magnetic field.
We are asked to find the inclination of the plane when it comes to equilibrium. We know that the condition for equilibrium is that the net force acting on it should be zero
Here since rotation is involved we can say the net torque should be zero at equilibrium.
Torque in the case of a current carrying coil is related to magnetic moment M and magnetic field B as
$\tau = M \times B = MB\sin \theta $
Magnetic moment of current carrying coil is given as
$M = NiA$
Where $N$ is the number of turns,$i$ is current through the coil and $A$ is the area of cross section.
Therefore we can write
$\tau = NiAB\sin \theta $
For torque to be zero the sin$\theta $ should be zero.
$\sin \theta $ will be zero when $\theta = {0^ \circ }$
Here we need to know that $\theta $ is the angle made by the area vector with the magnetic field.
Area vector is the normal vector to the surface of the area. That is, the angle between the perpendicular
to the plane and magnetic field should be zero. Which means the area vector will be along the direction of the field.
Therefore, the angle between plane and magnetic field should be perpendicular to each other so that the area vector will be along the direction of the field.
So, The correct answer is option C.
Note: Remember that the angle $\theta $ that appears in the equation for torque is the angle between area vector and the magnetic field. In the question we are asked to find the angle made by the plane with the magnetic field. Since the area vector is always directed perpendicular to the plane, if the angle made by it is ${0^ \circ }$ then the plane will be making an angle ${90^ \circ }$ to the field.
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