The magnetic field vector of an electromagnet | vedclass

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Question

(C) The magnetic force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

(C) The magnetic force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.At $t = 0$,the magnetic field is $\vec{B} = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz)$.For charge $q_1 = 4\pi$ at $z = \pi/k$:$\vec{B}_1 = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(k \cdot \frac{\pi}{k}) = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(\pi) = -B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.$\vec{F}_1 = q_1(\vec{v} 	imes \vec{B}_1) = 4\pi (0.5c\hat{i} 	imes (-B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}})) = -\frac{4\pi \cdot 0.5c B_0}{\sqrt{2}} (\hat{i} 	imes \hat{j}) = -\frac{2\pi c B_0}{\sqrt{2}} \hat{k}$.For charge $q_2 = 2\pi$ at $z = 3\pi/k$:$\vec{B}_2 = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(k \cdot \frac{3\pi}{k}) = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(3\pi) = -B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.$\vec{F}_2 = q_2(\vec{v} 	imes \vec{B}_2) = 2\pi (0.5c\hat{i} 	imes (-B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}})) = -\frac{2\pi \cdot 0.5c B_0}{\sqrt{2}} (\hat{i} 	imes \hat{j}) = -\frac{\pi c B_0}{\sqrt{2}} \hat{k}$.The ratio of the magnitudes is $\frac{|F_1|}{|F_2|} = \frac{2\pi c B_0 / \sqrt{2}}{\pi c B_0 / \sqrt{2}} = 2 : 1$.

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