A coil with 150 turns and a cross-sectional area of 1.00 m2 experiences a magnetic field whose strength increases by 0.65T in 1.
1 answer:
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The Lorentz force acting on the proton is equal to:
where
q is the proton charge
v is the proton speed
B is the intensity of the magnetic field
is the angle between the direction of v and B
since the proton travels perpendicularly to the magnetic field, in this case
, so the Lorentz force in this case is simply
The magnetic force provides also the centripetal force that keeps the proton in circular motion:
where
m is the proton mas
r is the radius of the orbit
If we re-arrange this equation and we use the data of the problem, we can find the radius of the orbit:
where
q is the proton charge
v is the proton speed
B is the intensity of the magnetic field
since the proton travels perpendicularly to the magnetic field, in this case
The magnetic force provides also the centripetal force that keeps the proton in circular motion:
where
m is the proton mas
r is the radius of the orbit
If we re-arrange this equation and we use the data of the problem, we can find the radius of the orbit: