A current i flows through a wire in the posit | vedclass

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(B) The magnetic force on a current-carrying wire is given by $\overrightarrow{F} = i(\overrightarrow{l} 	imes \overrightarrow{B})$.Given $\overright

Vedclass · Accepted Answer

$\sqrt{2} B_0il$

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(B) The magnetic force on a current-carrying wire is given by $\overrightarrow{F} = i(\overrightarrow{l} 	imes \overrightarrow{B})$.Given $\overrightarrow{l} = l\hat{i}$ and $\overrightarrow{B} = B_0(\hat{i} + \hat{j} + \hat{k})$.Substituting these values: $\overrightarrow{F} = i[l\hat{i} 	imes B_0(\hat{i} + \hat{j} + \hat{k})] = B_0il[\hat{i} 	imes \hat{i} + \hat{i} 	imes \hat{j} + \hat{i} 	imes \hat{k}]$.Using cross product rules $\hat{i} 	imes \hat{i} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,and $\hat{i} 	imes \hat{k} = -\hat{j}$,we get:$\overrightarrow{F} = B_0il(0 + \hat{k} - \hat{j}) = B_0il(\hat{k} - \hat{j})$.The magnitude of the force is $F = |\overrightarrow{F}| = B_0il \sqrt{(1)^2 + (-1)^2} = \sqrt{2} B_0il$.

$A$ current $i$ flows through a wire in the positive $X$-direction. The magnetic field is $\overrightarrow{B} = B_0(\hat{i} + \hat{j} + \hat{k}) \ T$. What is the magnitude of the force acting on a wire segment of length $l$?

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