$A$ single turn current loop in the shape of a right-angled triangle with sides $5\,cm$,$12\,cm$,and $13\,cm$ 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 $..........$

$A$ rectangular coil $ABCD$ is hung from one side of a balance as shown in the figure. $A$ $500 \, g$ mass is added to the other arm to balance the weight of the coil. $A$ current of $9.8 \, A$ is passed through the coil and a constant magnetic field of $0.4 \, T$ acting inward is switched on such that only arm $CD$ of length $1.5 \, cm$ lies in the field. The additional mass $m$ that must be added to regain the balance is: (in $, g$)

$A$ rectangular loop of sides $25 \text{ cm}$ and $10 \text{ cm}$ carrying a current of $10 \text{ A}$ is placed with its longer side parallel to a long straight conductor $10 \text{ cm}$ apart carrying a current of $25 \text{ A}$. The net force on the loop is

$A$ current $i = 2 \, A$ flows through a wire frame as shown in the figure. The frame is a combination of two equilateral triangles $ACD$ and $CDE$ with side length $1 \, m$. It is placed in a uniform magnetic field $B = 4 \, T$ acting perpendicular to the plane of the frame. The magnitude of the net magnetic force acting on the frame is.......$N$.

$A$ conducting bar $PQ$ of length $l$ carrying current $I$ is suspended from a rigid support as shown in the figure. $A$ uniform magnetic field $B$ perpendicular to $PQ$ and directed away from the reader (into the plane) is applied. If the mass of the bar is $M$,find the tension in each string.

$A$ 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: