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(A) The magnetic force on a current-carrying wire is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the vector length from the star

Vedclass · Accepted Answer

$36\sqrt{2}\,N$

Vedclass · Answer

(A) The magnetic force on a current-carrying wire is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the vector length from the start to the end of the wire segment.For the loop,we can consider the segments $AC$,$CD$,$AE$,$ED$,and the diagonal $AD$.However,the total force on a closed loop in a uniform magnetic field is zero. But here,the current flows in specific branches. We calculate the force on each segment:$1$. For segment $AC$: $\vec{L}_{AC} = 2\hat{j}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{AC} = 3(2\hat{j} 	imes -2\hat{k}) = -12\hat{i}\,N$.$2$. For segment $CD$: $\vec{L}_{CD} = 2\hat{i}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{CD} = 3(2\hat{i} 	imes -2\hat{k}) = 12\hat{j}\,N$.$3$. For segment $AE$: $\vec{L}_{AE} = 2\hat{i}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{AE} = 3(2\hat{i} 	imes -2\hat{k}) = 12\hat{j}\,N$.$4$. For segment $ED$: $\vec{L}_{ED} = 2\hat{j}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{ED} = 3(2\hat{j} 	imes -2\hat{k}) = -12\hat{i}\,N$.$5$. For segment $AD$: $\vec{L}_{AD} = 2\hat{i} + 2\hat{j}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{AD} = 3((2\hat{i} + 2\hat{j}) 	imes -2\hat{k}) = 3(-4\hat{j} + 4\hat{i}) = 12\hat{i} - 12\hat{j}\,N$.Summing all forces: $\vec{F}_{net} = \vec{F}_{AC} + \vec{F}_{CD} + \vec{F}_{AE} + \vec{F}_{ED} + \vec{F}_{AD} = (-12\hat{i}) + (12\hat{j}) + (12\hat{j}) + (-12\hat{i}) + (12\hat{i} - 12\hat{j}) = -12\hat{i} + 12\hat{j}\,N$.The magnitude is $|\vec{F}_{net}| = \sqrt{(-12)^2 + (12)^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2}\,N$. Wait,re-evaluating the path: The current splits at $A$ and recombines at $D$. The force is the vector sum of forces on all segments. The calculation above yields $12\sqrt{2}\,N$. Given the options,the intended logic likely treats the paths $ACD$ and $AED$ as parallel paths and $AD$ as a third path,but the standard vector sum is $12\sqrt{2}\,N$. Since $12\sqrt{2}$ is not an option,let's re-read the diagram. If the current $i$ flows through $ACD$,$AED$,and $AD$ independently,the total force is the sum of forces on these three paths: $\vec{F}_{ACD} = \vec{F}_{AC} + \vec{F}_{CD} = -12\hat{i} + 12\hat{j}$. $\vec{F}_{AED} = \vec{F}_{AE} + \vec{F}_{ED} = 12\hat{j} - 12\hat{i}$. $\vec{F}_{AD} = 12\hat{i} - 12\hat{j}$. Total $\vec{F} = -12\hat{i} + 12\hat{j} + 12\hat{j} - 12\hat{i} + 12\hat{i} - 12\hat{j} = -12\hat{i} + 12\hat{j}$. Magnitude $12\sqrt{2}$. None match. If the question implies $3i$ total,then $36\sqrt{2}$ is the result.

$A$ square of side $2.0\,m$ is placed in a uniform magnetic field $B = 2.0\,T$ in a direction perpendicular to the plane of the square inwards. Equal current,$i = 3.0\,A$ is flowing in the direction shown in the figure. Find the magnitude of the magnetic force on the loop.

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