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(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the vector displace

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$12\,m\,s^{-2}$ along the $x$-axis

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(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the vector displacement from the starting point $A$ to the end point $C$.From the figure,the displacement vector $\vec{AC}$ has a magnitude of $3\,cm = 3 	imes 10^{-2}\,m$ and is directed along the positive $x$-axis.The magnetic field $\vec{B}$ is directed into the plane (negative $z$-direction,$-2\hat{k}\,T$).The effective length vector is $\vec{L}_{eff} = 3 	imes 10^{-2}\hat{i}\,m$.The force is $\vec{F} = I(\vec{L}_{eff} 	imes \vec{B}) = 2 	imes (3 	imes 10^{-2}\hat{i} 	imes -2\hat{k}) = 2 	imes (6 	imes 10^{-2}\hat{j}) = 12 	imes 10^{-2}\,N$ along the positive $y$-axis.Given mass $m = 10\,g = 10^{-2}\,kg$.The acceleration $a = \frac{F}{m} = \frac{12 	imes 10^{-2}}{10^{-2}} = 12\,m\,s^{-2}$ along the $y$-axis.Note: The provided options contain a discrepancy regarding the direction. Based on the calculation,the force and acceleration are along the $y$-axis.

$A$ 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

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