In the given figure,the magnetic force on the | vedclass

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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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$20 \, N$

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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 start point $A$ to the end point $C$.The wire consists of a straight segment $AB$ of length $3 \, m$ and a curved segment $BC$ which is a quarter circle of radius $R = 4 \, m$.The coordinates are: $A = (0, 0)$,$B = (3, 0)$,and $C = (3, 4)$.The effective length vector $\vec{L}_{eff} = \vec{AC} = (3 \hat{i} + 4 \hat{j}) \, m$.The magnetic field is directed into the page,so $\vec{B} = -2 \hat{k} \, T$.The force is $\vec{F} = I(\vec{L}_{eff} 	imes \vec{B}) = 2 \, A 	imes [(3 \hat{i} + 4 \hat{j}) 	imes (-2 \hat{k})] \, T$.Using the cross product rules ($\hat{i} 	imes \hat{k} = -\hat{j}$ and $\hat{j} 	imes \hat{k} = \hat{i}$):$\vec{F} = 2 	imes [3(\hat{i} 	imes -2\hat{k}) + 4(\hat{j} 	imes -2\hat{k})] = 2 	imes [-6(-\hat{j}) + -8(\hat{i})] = 2 	imes [6\hat{j} - 8\hat{i}] = (12\hat{j} - 16\hat{i}) \, N$.The magnitude of the force is $F = \sqrt{(12)^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, N$.

In the given figure,the magnetic force on the wire $ABC$ will be $(B = 2 \, T, I = 2 \, A)$.

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