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Average velocity Of the particle in a large time interval.

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In a region of space, a uniform static magnetic field of induction $B_1$ is established above the x-z plane and another uniform static magnetic field of induction $B_2 \gt B_1$ is established below the x-z plane. Both fields are in the positive z-direction.
A particle of mass m and charge q is projected from the origin with velocity v making angle $\theta$ with x-axis as shown in the figure.
Find the average velocity of the particle in a large time interval.

But the answer is given as :

asked Aug 9, 2018 in Physics Problems by koolman (4,236 points)
edited Aug 9, 2018 by sammy gerbil
Your method is correct. Your mistakes (I think) are in the calculations of $T_1, T_2$ and $x$. Your formula for $T_2$ depends on $\theta$ but that for $T_1$ does not.  Why not? Likewise your formula for $x$ appears to be $2R_1-2R_2\sin\theta$. The 2nd term depends on $\theta$ but the 1st term does not. Why not?

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Your method of solution is correct. Your mistakes (I think) are in the calculations of $T_1, T_2$ and $x$, all of which should depend on angle $\theta$. Your formula for $T_2$ depends on $\theta$ but that for $T_1$ does not. Likewise your formula for $x$ appears to be $2R_1-2R_2\sin\theta$. The 2nd term depends on $\theta$ but the 1st term does not.


The motion in each region is a circular arc of radius $r$. The centripetal force for this motion is provided by the magnetic force on the particle : $$\frac{mv^2}{r}=Bqv$$ $$r=\frac{mv}{qB}$$ The distance moved along the $x$ axis is the chord of the arc, which is $2r\sin\theta$ where $\theta$ is the angle between the arc and the $x$ axis.

After tracing an arc in both regions, the motion repeats. During each cycle the resultant distance moved in the $+x$ direction is $$x=2(r_1-r_2)\sin\theta=\frac{2mv\sin\theta}{q}(\frac{1}{B_1}-\frac{1}{B_2})=\frac{2mv\sin\theta (B_2-B_1)}{qB_1B_2}$$

The angle subtended by each arc is $\theta_1=2\theta$ and $\theta_2=2(\pi-\theta)$ respectively. The times for the particle to traverse each arc are $\frac{r_1\theta_1}{v}, \frac{r_2\theta_2}{v}$ respectively. The total time for each cycle of the motion is therefore $$t=\frac{1}{v}(r_1\theta_1+r_2\theta_2)=\frac{1}{v}(2\theta\frac{mv}{qB_1}+2(\pi-\theta)\frac{mv}{qB_2})=\frac{2m((\pi-\theta) B_1+\theta B_2)}{qB_1B_2}$$

The average velocity of the particle in the $+x$ direction is $$V=\frac{x}{t}=\frac{(B_2-B_1)v\sin\theta}{(\pi-\theta)B_1+\theta B_2}=\frac{(B_2-B_1)v\sin\theta}{\pi B_1+\theta (B_2-B_1)}$$

answered Aug 13, 2018 by sammy gerbil (27,556 points)
selected Aug 22, 2018 by koolman
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