Answer

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**Hint:**We have a current carrying loop of radius R in a uniform magnetic field. The angle between the normal and the magnetic field is given to us. We know that a current carrying loop will only experience a torque in a uniform magnetic field.

**Complete step by step answer:**

In the question we are given a current carrying loop of radius R in a uniform magnetic field. The strength of the magnetic field is given as B. the angle between the field and the normal is given as $30{}^\circ $.

The figure below shows a current carrying loop as said in the question.

Here we have a uniform magnetic field and a current carrying loop.

We know that a current carrying loop in a uniform magnetic field will only experience torque, i.e. it will not have any force.

Therefore we can say that a current carrying loop in a uniform magnetic field will have a force which is equal to zero.

$\therefore Force=0$

**Hence the correct answer is option A.**

**Note:**

As said before, a current carrying loop in a uniform magnetic field will experience a torque.

The equation for torque on a current carrying loop in uniform magnetic field is given as,

$\tau =\vec{\mu }\times \vec{B}$, were ‘$\mu $’ is the magnetic moment and ‘B’ is the magnetic field.

$\Rightarrow \tau =\vec{\mu }\vec{B}\sin \theta $

Magnetic moment can be found using the equation,

$\mu =IA$, were ‘I’ is the current and ‘A’ is the area.

Since ‘R’ is the radius of the loop, we will get the area of the loop as,

$A=\pi {{R}^{2}}$

Therefore we get the magnetic moment as,

$\therefore \mu =\pi {{R}^{2}}I$

Therefore, by substituting this, the equation for torque will become,

$\therefore \tau =\left( \pi {{R}^{2}}I \right)\vec{B}\sin \theta $

From the question we have $\theta =30{}^\circ $

$\therefore \tau =\left( \pi {{R}^{2}}I \right)\vec{B}\sin 30$

$\Rightarrow \tau =\left( \pi {{R}^{2}}I \right)\vec{B}\times \dfrac{1}{2}$

$\therefore \tau =\dfrac{\pi }{2}\vec{B}{{R}^{2}}I$

This is the equation for torque of a current carrying loop in a uniform magnetic field.

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