At $t = 0$ a charge $q$ is at the origin and moving in the $y-$ direction with velocity $\overrightarrow v = v\,\hat j .$ The charge moves in a magnetic field that is for $y > 0$ out of page and given by $B_1 \hat z$ and for $y < 0$ into the page and given $-B_2 \hat z .$ The charge's subsequent trajectory is shown in the sketch. From this information, we can deduce that

Q: At $t = 0$ a charge $q$ is at the origin and moving in the $y-$ direction with velocity $\overrightarrow v = v\,\hat j .$ The charge moves in a magnetic field that is for $y > 0$ out of page and given by $B_1 \hat z$ and for $y < 0$ into the page and given $-B_2 \hat z .$ The charge's subsequent trajectory is shown in the sketch. From this information, we can deduce that

for first half: $\begin{gathered} \,\,{{\vec f}_m} = q(\vec V \times {{\vec B}_1}) \hfill \\ \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \hfill \\ \,\,\,\,i\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\hat j{\kern 1pt} \, \times \,\,( - k) \hfill \\ \end{gathered} $ $\therefore $ $q$ should be positive $\left|B_{0}\right|>\left|B_{1}\right|$ since, last tranjection is due to $\mathrm{B}_{2}-\mathrm{B}_{1}$ and decides by $\mathrm{B}_{2}$

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Solution

for first half: $\begin{gathered}
\,\,{{\vec f}_m} = q(\vec V \times {{\vec B}_1}) \hfill \\
\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \hfill \\
\,\,\,\,i\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\hat j{\kern 1pt} \, \times \,\,( - k) \hfill \\
\end{gathered} $

$\therefore $ $q$ should be positive $\left|B_{0}\right|>\left|B_{1}\right|$ since, last tranjection is due to $\mathrm{B}_{2}-\mathrm{B}_{1}$ and decides by $\mathrm{B}_{2}$

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