A conducting wire bent in the form of a parab | vedclass

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(D) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the

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$-96 \hat{i} \ N$

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(D) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the vector displacement from the starting point to the end point of the wire.From the parabola equation $y^2 = 2x$,at $x = 2$,$y^2 = 4$,so $y = \pm 2$. The points are $A(2, 2)$ and $B(2, -2)$.The current flows from $A$ to $B$ through the origin $O$. Thus,the effective length vector $\vec{L}$ is the vector from $A$ to $B$,which is $\vec{L} = (2-2)\hat{i} + (-2-2)\hat{j} = -4\hat{j} \ m$.Given $I = 4 \ A$ and $\vec{B} = 6\hat{k} \ T$.Substituting these values: $\vec{F} = 4 	imes (-4\hat{j} 	imes 6\hat{k}) = 4 	imes (-24)(\hat{j} 	imes \hat{k}) = -96\hat{i} \ N$.

$A$ 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

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