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Question

(C) The magnetic force on a current-carrying wire is given by $\vec F = I \int d\vec l 	imes \vec B$. Consider a square loop with vertices at $(x, 0)

Vedclass · Accepted Answer

$I_0 B_0 l$

Vedclass · Answer

(C) The magnetic force on a current-carrying wire is given by $\vec F = I \int d\vec l 	imes \vec B$. Consider a square loop with vertices at $(x, 0)$,$(x+l, 0)$,$(x+l, l)$,and $(x, l)$. For the two vertical segments (parallel to the $y$-axis): $1$. At $x_1 = x$,$\vec B_1 = B_0 [1 + \frac{x}{l}] \hat k$. The force is $\vec F_1 = I_0 (l \hat j) 	imes (B_0 [1 + \frac{x}{l}] \hat k) = I_0 l B_0 [1 + \frac{x}{l}] \hat i$. $2$. At $x_2 = x+l$,$\vec B_2 = B_0 [1 + \frac{x+l}{l}] \hat k = B_0 [2 + \frac{x}{l}] \hat k$. The force is $\vec F_2 = I_0 (-l \hat j) 	imes (B_0 [2 + \frac{x}{l}] \hat k) = -I_0 l B_0 [2 + \frac{x}{l}] \hat i$. For the two horizontal segments (parallel to the $x$-axis): The forces on the top and bottom segments are equal and opposite because the magnetic field depends only on $x$,so they cancel out. The net force is $\vec F_{net} = \vec F_1 + \vec F_2 = I_0 l B_0 [1 + \frac{x}{l} - 2 - \frac{x}{l}] \hat i = -I_0 l B_0 \hat i$. The magnitude of the net magnetic force is $|\vec F_{net}| = I_0 B_0 l$.

The magnetic field existing in a region is given by $\vec B = B_0 [1 + \frac{x}{l}] \hat k$. $A$ square loop of edge $l$ carrying current $I_0$ is placed with its edges parallel to the $x-y$ axes. Find the magnitude of the net magnetic force experienced by the loop.

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