$A$ square loop of side $a$ hangs from an insulating hanger of a spring balance. The magnetic field of strength $B$ exists only at the lower edge. It carries a current $I$. Find the change in the reading of the spring balance if the direction of current is reversed.

$A$ loop of flexible conducting wire lies in a magnetic field of $2.0 \,T$ with its plane perpendicular to the field. The length of the wire is $1 \,m$. When a current of $1.1 \,A$ is passed through the loop, it opens into a circle. The tension developed in the wire is (in $\,N$)

An infinitely long current-carrying wire and a small current-carrying loop are in the plane of the paper as shown. The radius of the loop is $a$ and the distance of its centre from the wire is $d$ $(d >> a)$. If the loop applies a force $F$ on the wire,then:

The figure shows a conducting loop $ABCDA$ placed in a uniform magnetic field perpendicular to its plane. The part $ABC$ is the $(3/4)^{th}$ portion of the square of side length $l$. The part $ADC$ is a circular arc of radius $R$. The points $A$ and $C$ are connected to a battery which supplies a current $I$ to the circuit. The magnetic force on the loop due to the field $B$ is

$A$ straight wire is placed in a magnetic field that varies with distance $x$ from the origin as $\vec{B} = B_0 \left( 2 - \frac{x}{a} \right) \hat{k}$. The ends of the wire are at $(a, 0)$ and $(2a, 0)$ and it carries a current $I$ in the positive $x$-direction. If the force on the wire is $\vec{F} = IB_0 \left( \frac{ka}{2} \right) \hat{j}$, then the value of $k$ is:

$AB$ and $CD$ are smooth parallel rails,separated by a distance $l$,and inclined to the horizontal at an angle $\theta$. $A$ uniform magnetic field of magnitude $B$,directed vertically upwards,exists in the region. $EF$ is a conductor of mass $m$,carrying a current $i$. If the conductor is in equilibrium,find the relation between the given parameters.