Girra Past Paper q 2020 (1 Viewer)

hmim

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1685103446394.png ans

How is the answer to this (A)? How would i go about doing it?
 

carrotsss

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This gives me flashbacks I got so confused by this question, and my teacher couldnt even figure it out at first

First, the simple part. For the object to stay in circular motion, moved solely by the magnetic field, the centripetal force must be equal to the magnetic force.
So, qvB=mv^2/r
Rearranging: B=mv/qr
But, if you just substitute these values into the equation, you’ll get 0.0159, which isn’t an answer. The reason for this is that the objects velocity is close enough to the speed of light that we have to consider special relativity, so you have to consider the length contraction which occurs to the radius using the formula on ur ref sheet (you might not have learnt this yet), and you should get r=74.55m using that. Then, you can sub it into the formula for magnetic field and find the answer to be 0.0175T
 
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wizzkids

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Have a look at that speed ... 1.25 x 108 m/s is 41.6 percent of the speed of light.
That should alert you straight away. You are going to need the mass dilation equation. By the way, the radius of the proton path is measured in the stationary frame of reference of the observer and the synchrotron, so it is true length. There is no length contraction effect from the frame of reference of the observer.
B=mv/qr is the correct equation to use, but you must substitute the dilated mass of the proton, which at 41.6 percent of the speed of light turns out to be 1.100 times its rest mass. After substitution into the equation, the B-field is 0.0175 tesla.
 
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