A uniform magnetic field vec B = (3hat i + 4h | vedclass

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

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$\sqrt{2} (\hat i - \hat j + \hat k) \, N$

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(B) The magnetic force on a current-carrying wire is given by $\vec F = i (\vec L 	imes \vec B)$,where $\vec L$ is the displacement vector from the start to the end of the wire.From the figure,the wire is a semicircle of radius $R = 1 \, m$ oriented at $45^\circ$ to the $x$-axis.The start point is $(2 - R \cos 45^\circ, 2 - R \sin 45^\circ, 0) = (2 - 1/\sqrt{2}, 2 - 1/\sqrt{2}, 0)$.The end point is $(2 + R \cos 45^\circ, 2 + R \sin 45^\circ, 0) = (2 + 1/\sqrt{2}, 2 + 1/\sqrt{2}, 0)$.The displacement vector $\vec L = (x_f - x_i)\hat i + (y_f - y_i)\hat j = (2/\sqrt{2})\hat i + (2/\sqrt{2})\hat j = \sqrt{2}\hat i + \sqrt{2}\hat j = \sqrt{2}(\hat i + \hat j)$.Given $\vec B = 3\hat i + 4\hat j + \hat k$ and $i = 1 \, A$.$\vec F = 1 	imes [\sqrt{2}(\hat i + \hat j) 	imes (3\hat i + 4\hat j + \hat k)]$.$\vec F = \sqrt{2} [(\hat i 	imes 3\hat i) + (\hat i 	imes 4\hat j) + (\hat i 	imes \hat k) + (\hat j 	imes 3\hat i) + (\hat j 	imes 4\hat j) + (\hat j 	imes \hat k)]$.Using cross products: $\hat i 	imes \hat i = 0, \hat i 	imes \hat j = \hat k, \hat i 	imes \hat k = -\hat j, \hat j 	imes \hat i = -\hat k, \hat j 	imes \hat j = 0, \hat j 	imes \hat k = \hat i$.$\vec F = \sqrt{2} [0 + 4\hat k - \hat j - 3\hat k + 0 + \hat i] = \sqrt{2}(\hat i - \hat j + \hat k) \, N$.

$A$ uniform magnetic field $\vec B = (3\hat i + 4\hat j + \hat k) \, T$ exists in a region of space. $A$ semicircular wire of radius $1 \, m$ carrying a current of $1 \, A$ has its centre at $(2, 2, 0)$ and is placed in the $x-y$ plane as shown in the figure. The force on the semicircular wire is:

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