The figure shows a conducting loop $ADCA$ carrying current $i$ and placed in a region of uniform magnetic field $B_0$. The part $ADC$ forms a semicircle of radius $R$. The magnitude of the force on the semicircular part of the loop is equal to

Write the magnetic force equation on a current-carrying element $I\vec{dl}$ inside a magnetic field $\vec{B}$. Write the law used to determine the direction of the magnetic force.

$A$ 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:

As shown in the figure, a wire is bent to form a $D$-shaped loop carrying current $I$, where the curved part is a semi-circle of radius $R$. The loop is placed in a uniform magnetic field $\overrightarrow{B}$, which is directed into the plane of the paper. The net magnetic force on the closed loop is:

The circuit in the figure consists of wires at the top and bottom and identical springs as the left and right sides. The wire at the bottom has a mass of $10 \, g$ and is $5 \, cm$ long. The wire is hanging as shown in the figure. The springs stretch $0.5 \, cm$ under the weight of the wire and the circuit has a total resistance of $12 \, \Omega$. When the lower wire is subjected to a static magnetic field,the springs stretch an additional $0.3 \, cm$. The magnetic field is

In the given diagram, the loop tends to: