The following figure shows the path of an ele | vedclass

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(A) $1$. For an electron (negatively charged particle) moving with velocity $v$ in a magnetic field $B$,the magnetic force is given by $F = q(v 	imes

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$B_1$ is into the page and it is stronger than $B_2$

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(A) $1$. For an electron (negatively charged particle) moving with velocity $v$ in a magnetic field $B$,the magnetic force is given by $F = q(v 	imes B)$. Since the charge $q$ is negative,the force is in the direction opposite to $(v 	imes B)$.$2$. Using Fleming's Left-Hand Rule for a positive charge,the force is perpendicular to both $v$ and $B$. For an electron,we use the Right-Hand Rule or reverse the direction obtained from the Left-Hand Rule.$3$. In region $B_1$,the electron enters and turns clockwise. By applying the right-hand rule for a negative charge,the magnetic field $B_1$ must be directed into the page.$4$. In region $B_2$,the electron turns counter-clockwise. By applying the same rule,the magnetic field $B_2$ must be directed out of the page.$5$. The radius of the circular path is given by $r = \frac{mv}{qB}$. Since $m, v,$ and $q$ are constant,$r \propto \frac{1}{B}$.$6$. From the figure,the radius of the path in region $B_1$ $(r_1)$ is smaller than the radius of the path in region $B_2$ $(r_2)$.$7$. Since $r_1  B_2$. Thus,$B_1$ is stronger than $B_2$.

The following figure shows the path of an electron that passes through two regions containing uniform magnetic fields of magnitudes $B_1$ and $B_2$. Its path in each region is a half-circle. Choose the correct option.

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