A proton moving at 3.0x10^7 m/s experiences a magnetic field of 4.0T.  Calculate the radius of the path of the proton    



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Posted on (Answer #1)

You may evaluate the radius of curvature of the path using the following formula, such that:

`r = (m*v)/(q*B)`

The problem provides `v = 3*10^7 m/s` and `B = 4.0T` , hence, considering `q = 1.6*10^(-19)C` and `m = 9.11*10^(-31)Kg` , yields:

`r = (9.11*10^(-31)*3*10^7)/(1.6*10^(-19)*4)`

`r = 4.27*10^(-5) m`

Hence, evaluating the radius of curvature, under the given conditions, yields `r = 4.27*10^(-5) m.`

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