$A$ current $i$ flows through a wire in the positive $X$-direction. The magnetic field is $\overrightarrow{B} = B_0(\hat{i} + \hat{j} + \hat{k}) \ T$. What is the magnitude of the force acting on a wire segment of length $l$?

The net charge in a current-carrying wire is zero,yet a magnetic field exerts a force on it. This is because a magnetic field exerts a force on:

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.)

Three long,straight parallel wires carrying currents are arranged as shown. The wire $C$ which carries a current of $5.0 \text{ A}$ is so placed that it experiences no force. The distance of wire $C$ from wire $D$ is (in $cm$)

$A$ uniform conducting wire $ABC$ has a mass of $10\,g$. $A$ current of $2\,A$ flows through it. The wire is kept in a uniform magnetic field $B = 2\,T$ directed into the plane of the paper. The acceleration of the wire will be

$A$ 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