$A$ $2 \, A$ current carrying straight metal wire of resistance $1 \, \Omega$, resistivity $2 \times 10^{-6} \, \Omega m$, area of cross-section $10 \, mm^2$ and mass $500 \, g$ is suspended horizontally in mid-air by applying a uniform magnetic field $\vec{B}$. The magnitude of $B$ is . . . . . . . $\times 10^{-1} \, T$ (given, $g=10 \, m/s^2$).

$A$ massless square loop of wire of resistance $R$ supporting a mass $M$ hangs vertically with one of its sides in a uniform magnetic field $B$ directed outwards in the shaded region. $A$ d.c. voltage $V$ is applied to the loop. For what value of $V$ will the magnetic force exactly balance the weight of the supporting mass $M$? (side of loop $= L, g =$ acceleration due to gravity)

$A$ wire of length $L$ carries a current $I$ along the $X$-axis. The magnetic force acting on the wire is given by $\vec{F} = I B_0 L(\hat{k} - \hat{j})$. The existing magnetic field $\vec{B}$ is

$A$ straight wire of length $2 \ m$ carries a current of $10 \ A$. If this wire is placed in a uniform magnetic field of $0.15 \ T$ making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be

$A$ conducting wire $PQ$ carries a current $10 \ A$ as shown in the figure. It is placed in a uniform magnetic field $5 \ T$ which is acting normally outwards from the paper. The net force experienced by the wire is:

The magnetic field existing in a region is given by $\vec{B} = 0.2(1 + 2x) \hat{k} \text{ T}$. $A$ square loop of edge $50 \text{ cm}$ carrying $0.5 \text{ A}$ current is placed in the $x-y$ plane with its edges parallel to the $x-y$ axes,as shown in the figure. The magnitude of the net magnetic force experienced by the loop is . . . . . . $\text{mN}$.