The force on a current-carrying loop (radius $= R$) in a uniform magnetic field $(B)$,which is at an angle $30^{\circ}$ with the normal,will be:

In the given figure,the magnetic force on the wire $ABC$ will be $(B = 2 \, T, I = 2 \, A)$.

$A$ current $i$ flows through a wire in the positive $X$-direction. The magnetic field is $\overrightarrow{B} = B_0(\hat{i} + \hat{j} + \hat{k}) \ T$. What is the magnitude of the force acting on a wire segment of length $l$?

Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They exert

Obtain the expression for the force between two parallel current-carrying wires.

$A$ single-turn current loop in the shape of a right-angled triangle with sides $5 \,cm, 12 \,cm, 13 \,cm$ carries a current of $2 \,A$. The loop is placed in a uniform magnetic field of magnitude $0.75 \,T$, whose direction is parallel to the current in the $13 \,cm$ side of the loop. The magnitude of the magnetic force on the $5 \,cm$ side is $\frac{x}{130} \,N$. The value of '$x$' is: