At t = 0,a charge q is at the origin and movi | vedclass

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(A) $1$. For $y > 0$,the magnetic field is $\vec{B}_1 = B_1\hat{k}$ (out of the page). The velocity is $\vec{v} = v\hat{j}$. The magnetic force is $\v

Vedclass · Accepted Answer

$q > 0$ and $|B_1| < |B_2|$

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(A) $1$. For $y > 0$,the magnetic field is $\vec{B}_1 = B_1\hat{k}$ (out of the page). The velocity is $\vec{v} = v\hat{j}$. The magnetic force is $\vec{F}_m = q(\vec{v} 	imes \vec{B}_1) = q(v\hat{j} 	imes B_1\hat{k}) = qvB_1\hat{i}$.$2$. From the diagram,the particle curves towards the positive $x$-axis for $y > 0$,meaning the force is in the $+\hat{i}$ direction. Since $v, B_1 > 0$,$q$ must be positive $(q > 0)$.$3$. The radius of the circular path is given by $R = \frac{mv}{qB}$. The radius $R_1$ in the region $y > 0$ is $R_1 = \frac{mv}{qB_1}$ and the radius $R_2$ in the region $y $4$. Looking at the trajectory,the path in the $y  0$ region,implying $R_2 $5$. Since $R_2  B_1$,or $|B_1| < |B_2|$.

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