Uniform magnetic fields of different strength | vedclass

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Question

(D) Since the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the charged particle moves in a circular path with radius $R = \frac

Vedclass · Accepted Answer

$\frac{mv}{qB_1}\left(1-\frac{B_1}{B_2}ight) 	imes 2$

Vedclass · Answer

(D) Since the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the charged particle moves in a circular path with radius $R = \frac{mv}{qB}$.In region $2$,the particle completes a semi-circle of radius $R_2 = \frac{mv}{qB_2}$ and returns to the interface at point $D$. The distance $AD = 2R_2 = \frac{2mv}{qB_2}$.Then,the particle enters region $1$ and completes a semi-circle of radius $R_1 = \frac{mv}{qB_1}$ and returns to the interface at point $C$. The distance $AC = 2R_1 = \frac{2mv}{qB_1}$.The net displacement along the interface is the distance between the starting point $A$ and the final point $C$. Since the particle moves in opposite directions in the two regions relative to the interface,the displacement is $AC = AD - CD$ is incorrect based on the geometry. Looking at the path,the particle moves from $A$ to $D$ (in region $2$) and then from $D$ to $C$ (in region $1$).The total displacement along the interface is $AC = AD - CD$ is not correct; rather,the particle moves from $A$ to $D$ and then from $D$ to $C$. The distance $AC = AD - CD$ is not the correct interpretation. The distance $AC = 2R_1 - 2R_2 = \frac{2mv}{qB_1} - \frac{2mv}{qB_2} = \frac{2mv}{q} \left(\frac{1}{B_1} - \frac{1}{B_2}\right) = \frac{2mv}{qB_1} \left(1 - \frac{B_1}{B_2}\right)$.

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