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(A) The magnetic force on a current-carrying wire segment of length vector $\vec{L}$ in a uniform magnetic field $\vec{B}$ is given by $\vec{F} = i(\v

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$ilB$

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(A) The magnetic force on a current-carrying wire segment of length vector $\vec{L}$ in a uniform magnetic field $\vec{B}$ is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$.For a closed loop in a uniform magnetic field,the net force is $\vec{F}_{net} = i(\oint d\vec{l}) 	imes \vec{B}$. Since the integral of $d\vec{l}$ over a closed loop is zero,the net force on any closed loop in a uniform magnetic field is zero.However,in this problem,the current enters at $C$ and leaves at $A$. The triangle is not a closed loop for the current flow. We can treat this as three segments: $CB$,$BA$,and $AC$.Using the principle of superposition,the net force is the vector sum of forces on each segment: $\vec{F}_{net} = \vec{F}_{CB} + \vec{F}_{BA} + \vec{F}_{AC}$.Each segment has length $l$ and is perpendicular to the field $B$. The force on each segment is $F = ilB$ directed radially outward or inward depending on the current direction.By symmetry and vector addition,the resultant force on the segments $CB$ and $BA$ effectively acts as a force on the displacement vector from $C$ to $A$,which is $l$. Thus,the net force is equivalent to the force on a straight wire of length $l$ carrying current $i$ from $C$ to $A$,which is $ilB$.

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:

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