An isosceles triangular current-carrying loop is placed in a uniform magnetic field $\overrightarrow{B_{o}}$ directed perpendicular to the plane of the loop,as shown in the figure. The net magnetic force on the loop is:

Two infinitely long parallel conducting wires $A$ and $B$ carry currents $I$ and $2I$, respectively, in the same direction. The wire $A$ has uniform mass per unit length $\lambda$ and lies on an insulated floor. The wire $B$ is kept fixed at a height $h$ above the floor. The minimum magnitude of $h$ so that the wire $A$ does not rise from the floor is : [$g$ is the acceleration due to gravity and $\mu_0$ is the permeability of free space]

The force per unit length on a straight wire carrying a current of $8 \ A$ making an angle of $30^{\circ}$ with a uniform magnetic field of $0.15 \ T$ is: (in $N \ m^{-1}$)

$A$ long straight wire is carrying current $I_1$ in the $+z$ direction. The $x-y$ plane contains a closed circular loop carrying current $I_2$ and does not encircle the straight wire. The net force on the loop will be:

The force acting per unit length on a very long straight conductor carrying a steady current of $1 \,A$ flowing from south to north is: (The horizontal component of the earth's magnetic field at the place is $3 \times 10^{-5} \,T$ and its direction is from geographical south to geographical north.)

The figure shows an equilateral triangle $ABC$ of side length '$l$' carrying currents as shown,placed in a uniform magnetic field '$B$' perpendicular to the plane of the triangle. The magnitude of the net magnetic force on the triangle is: