$A$ conductor in the form of a right angle $ABC$ with $AB = 3\, cm$ and $BC = 4\, cm$ carries a current of $10\, A$. There is a uniform magnetic field of $5\, T$ perpendicular to the plane of the conductor. The force on the conductor will be......$N$

$A$ wire is bent in the form of an equilateral triangle of side $100 \,cm$ and carries a current of $2 \,A$. It is placed in a magnetic field of induction $2.0 \,T$ directed perpendicular into the plane of paper. The direction and magnitude of magnetic force acting on each side of the triangle will be

$A$ semicircular ring of radius $R$ carrying current $i$ is placed in a uniform magnetic field of intensity $B$ such that the plane of the wire is perpendicular to the magnetic field,as shown in the figure. The net force acting on the ring is

$A$ metal ring of radius $r = 0.5 \, m$ with its plane normal to a uniform magnetic field $B$ of induction $0.2 \, T$ carries a current $I = 100 \, A$. The tension in newtons developed in the ring is:

$A$ straight wire of length $0.5 \ m$ and carrying a current of $1.2 \ A$ is placed in a uniform magnetic field of induction $2 \ T$. The magnetic field is perpendicular to the length of the wire. What is the force acting on the wire (in $N$)? $[\sin 90^{\circ} = 1]$

If a straight current-carrying wire of linear density $0.12 \ kg \ m^{-1}$ is suspended in mid-air by a uniform horizontal magnetic field of $0.5 \ T$ normal to the length of the wire,then the current through the wire is (Acceleration due to gravity $= 10 \ m \ s^{-2}$; Neglect earth's magnetic field) (in $A$)