The magnetic force on a current-carrying wire in an external uniform magnetic field depends on:

$A$ 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:

In the given figure,$X$ and $Y$ are two long straight parallel conductors each carrying a current of $2\,\text{A}$. The force on each conductor is $F$ newtons. When the current in each is changed to $1\,\text{A}$ and reversed in direction,the force on each is now

$A$ conducting wire bent in the form of a parabola $y^2=2x$ carries a current of $4 \ A$ as shown in the figure. This wire is placed in a uniform magnetic field of induction $\vec{B}=+6 \hat{k} \ T$. The magnetic force on the wire is

$A, B$ and $C$ are three parallel conductors of equal lengths carrying currents $I, I$ and $2I$ respectively. The distance between $A$ and $B$ is $x$ and that between $B$ and $C$ is also $x$. $F_1$ is the force exerted by conductor $B$ on $A$. $F_2$ is the force exerted by conductor $C$ on $A$. The current $I$ in $A$ and $I$ in $B$ are in the same direction,and the current $2I$ in $C$ is in the opposite direction. Then:

An elastic circular wire of length $l$ carries a current $I$. It is placed in a uniform magnetic field $\vec{B}$ (out of the paper) such that its plane is perpendicular to the direction of $\vec{B}$. The wire will experience: