Find Speed of Particles in B-Field

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To find the speed of particles entering a magnetic field (B-field), kinetic energy can be used, specifically 100 keV. The equation used is (1/2)mv² = U, where U is the kinetic energy. The calculation involves substituting the mass of a proton (1.67 x 10^-27 kg) into the equation to solve for speed. The formula derived is v = √(2 * 1.602 x 10^-14 / 1.67 x 10^-27). This approach is correct for determining the speed of the particles.
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given is:
http://img340.imageshack.us/img340/6684/oplosqa0.jpg

So if you have a question with a B-field you need any speed given from the particles who entering in this field.
Who can I find this speed? Thanks.
 
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Bert said:
given is:
http://img340.imageshack.us/img340/6684/oplosqa0.jpg

So if you have a question with a B-field you need any speed given from the particles who entering in this field.
Who can I find this speed? Thanks.

Get the speed from the kinetic energy 100keV.
 
Last edited by a moderator:
I try this but I fail so:

\frac{1}{2}m . v^2=U
then we have
1,602 \ . \ 10^{-19} \ . \ 1 . \ 10^5=1,602 \ . \ 10^{-14}= \frac{1}{2} 10^{-27}.v^2

so v must be \sqrt{ \frac{2 \ . \ 1,602 . 10^{-14} } {10 ^{-27} } }

is this correct?
 
Bert said:
I try this but I fail so:

\frac{1}{2}m . v^2=U
then we have
1,602 \ . \ 10^{-19} \ . \ 1 . \ 10^5=1,602 \ . \ 10^{-14}= \frac{1}{2} 10^{-27}.v^2

so v must be \sqrt{ \frac{2 \ . \ 1,602 . 10^{-14} } {10 ^{-27} } }

is this correct?

mass of a proton = 1.67*10^{-27}kg. Other than that everything looks good.
 
Thanks.
 
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