- #26

ehild

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x is correct. I meant z, sorry.Sorry . I think It should be ##x=\frac{v_1m}{qB}sinωt## only . I don't see why there should be a minus sign .

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- Thread starter Vibhor
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- #26

ehild

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x is correct. I meant z, sorry.Sorry . I think It should be ##x=\frac{v_1m}{qB}sinωt## only . I don't see why there should be a minus sign .

- #27

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##z = - \frac{v_1m}{qB}(1-cosωt) ## ??

- #28

ehild

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Yes.##z = - \frac{v_1m}{qB}(1-cosωt) ## ??

- #29

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x will be 0 when ##\frac{qBt}{\pi m }## is an integer , y will be 0 at ##t = \frac{2mv_2}{qE}## , z will be 0 when ##\frac{qBt}{2\pi m }## is an integer

- #30

ehild

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What integer is it? When is z = 0? And all of them should be zero at the same time.

x will be 0 when ##\frac{qBt}{\pi m }## is an integer , y will be 0 at ##t = \frac{2mv_2}{qE}## , z will be 0 when ##\frac{qBt}{2\pi m }## is an integer

- #31

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Option B)

- #32

ehild

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Correct . Was it difficult?Option B)

- #33

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Was it difficult?

It surely was until you took over . Honestly speaking , I had almost made up my mind to leave this problem as I was completely clueless . But last couple of hours of problem solving were pretty exciting ,thanks to you .

You are amazing .

- #34

ehild

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It was the combination of a circular motion in the x,z plane and a vertical projectile motion in the direction of y. The period of the circular motion should be the same as the flight time of the vertical projectile.

Thank you for the thanks

Thank you for the thanks

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