A circular flexible current loop of radius \[R\] carrying current \[i\] is placed in an inward magnetic field \[B\]. If we spin the loop with angular speed \[\omega \], then tension in string (Assume the mass of the loop is \[m\])

A. Is zero
B. Is more than \[iBR\]
C. Is less than \[iBR\]
D. Does not depend on rotation

Hint:To find the tension in the string. At first take a small section of the loop. Draw a diagram showing the tension and its component along x-axis and y-axis. Observe the diagram carefully, balance the required forces and take out the value of tension at equilibrium.

Complete step by step solution:
Given, radius of the loop is \[R\]
Current through the loop is \[i\]
Strength of magnetic field is \[B\]
Let \[T\] be the tension.
Let us take a small segment of the circular loop subtended angle \[2\theta \]. We draw a diagram representing the components of tension.


We resolve tension \[T\] into its component on both sides.
We observe that the components \[T\cos \theta \] on both sides are equal and in the opposite direction, so they get cancelled. So the remaining tension is,

\[{T_{net}} = 2T\sin \theta \] (i)

Since the loop is in a magnetic field, it will experience a magnetic force.
We have the formula for magnetic force on a current carrying wire due to a magnetic field as,
\[F = IlB\sin \theta \] (ii)
where \[I\] is the current flowing through the wire, \[l\] is the length, \[B\] is the magnetic field and
\[\theta \] is the angle between magnetic field and force.
Here, current \[I = i\]
Length of the small segment using the formula of arc is , \[l = R2\theta \]
And here \[\theta = {90^ \circ }\] so \[\sin \theta = 1\]

Using these values in equation (ii), we get the force as,
\[F = iR2\theta B\] (iii)

For equilibrium, the tension in the string should balance the force so, equating (iii) and (i) we get
\[2T\sin \theta = iR2\theta B \\
\Rightarrow T\sin \theta = iRB\theta \]
Since \[\theta \] is small, so we can write \[\sin \theta \simeq \theta \] then the equation
becomes,
\[T\theta = iRB\theta \\
\Rightarrow T = iRB \]
Therefore, for equilibrium tension is \[iRB\] , so to spin the loop with angular speed \[\omega \] the value of tension should be more than that of its equilibrium value that is more than \[iRB\].

Hence, the correct answer is option (B) is more than \[iRB\].

Note: In such types of questions, always take a small section of the given structure; draw a diagram showing all the forces and try to resolve the component of the forces. This will help in cancelling the equal and opposite forces and will make the calculations easy. To find the value of quantities in equilibrium, always balance the forces.