A equilateral triangular loop ADC of side $l$ carries a current $i$ in the direction as shown in the figure. The loop is kept in a uniform horizontal magnetic field $\vec B$ as shown in the figure. Then the net force on the loop is given as:

A. 0
B. $\dfrac{\sqrt3}{2}ilB$
C. perpendicular to paper inwards
D. perpendicular to paper outwards

Hint: We know that electricity and magnetism are inter connected. We know that a current carrying conductor produces a magnetic field around itself. And similarly, a varying magnetic field induces a current in the coil which is kept in the near surrounding.

Formula:
$\vec F=I\vec l\times \vec B$

Complete answer:
Since electricity and magnetism are interrelated, we know that a current carrying conductor when placed in a magnetic field experiences a force. Also, the direction force is given by the right hand rule.
The force $\vec F$ experienced by a current $I$ carrying coil of length $\vec l$ which is placed in a magnetic field $\vec B$ is given as $\vec F=I\vec l\times \vec B=Ilbsin\theta$, where $\theta$ is the angle made by the length vector with respect to the magnetic field.
The direction of the force is given by the right hand rule, where the thumb, index and the middle finger are extended outwards, such that they are perpendicular to each other. Then if the thumb points towards the direction of current, the index finger points towards the direction of the magnetic field, then the middle finger points the direction of the force.
Using this we can find the force on the given triangle. Consider the line segment AB, using the right hand rule, we get, that the direction of force is outward from the paper.
Let us now consider the line segment AC, using the right hand rule, we get that the direction of force is towards the paper.
Since the length, current and the magnetic field are the same, we can say the both cancel each other.
Now consider the line segment BC, since the $\theta=0\implies sin 0=0$, thus the force due to BC is also zero.
Hence the net force due to the given triangle is also zero.

Hence the correct option is a A $0$.

Note:
We know that, Force is a vector quantity, which has both magnitude and direction.Since the force is a cross product of both the line vector and the magnetic field, we can clearly say that the force is perpendicular to both the line vector and the magnetic field.