- #1
Amith2006
- 427
- 2
1) An electron and a proton are injected into a uniform magnetic field at right angles to the direction of the field with the same Kinetic Energy. Then:
a)the electron trajectory will be less curved than the proton trajectory
b)the proton trajectory will be less curved than the electron trajectory
c)both the trajectories will be equally curved
d)both the trajectories will be straight
My reasoning is as follows:
Let –e be the charge of electron, +e be the charge of proton and B be the uniform magnetic field. Let mass of proton be Mp and mass of electron be Me. Let v1 and v2 be the velocity of electron and proton respectively.
We know that Mp > Me
We know that a charged particle describes a circular path in a perpendicular magnetic field.
So, Bqv = (mv^2)/r
r = (mv^2)/(Bqv)
= {(1/2)mv^2}/{(1/2)Bqv}
Since kinetic energy, magnetic field and charge are same for electron and proton,
r proportional to 1/v
Now, K.E of electron = K.E of proton
(1/2)(Me)(v1^2) = (1/2)(Mp)(v2^2)
(v1/v2)^2 = Mp/Me
Since Mp > Me, v1>v2
Now, r1/r2 = v2/v1
Since v2<v1, r1<r2
So, radius of circular path of electron is less than radius of circular path of proton. So the electron path is more curved that proton path because curvature is the reciprocal of radius of curvature. But the book answer is a)the electron trajectory will be less curved than the proton trajectory. Please help!
a)the electron trajectory will be less curved than the proton trajectory
b)the proton trajectory will be less curved than the electron trajectory
c)both the trajectories will be equally curved
d)both the trajectories will be straight
My reasoning is as follows:
Let –e be the charge of electron, +e be the charge of proton and B be the uniform magnetic field. Let mass of proton be Mp and mass of electron be Me. Let v1 and v2 be the velocity of electron and proton respectively.
We know that Mp > Me
We know that a charged particle describes a circular path in a perpendicular magnetic field.
So, Bqv = (mv^2)/r
r = (mv^2)/(Bqv)
= {(1/2)mv^2}/{(1/2)Bqv}
Since kinetic energy, magnetic field and charge are same for electron and proton,
r proportional to 1/v
Now, K.E of electron = K.E of proton
(1/2)(Me)(v1^2) = (1/2)(Mp)(v2^2)
(v1/v2)^2 = Mp/Me
Since Mp > Me, v1>v2
Now, r1/r2 = v2/v1
Since v2<v1, r1<r2
So, radius of circular path of electron is less than radius of circular path of proton. So the electron path is more curved that proton path because curvature is the reciprocal of radius of curvature. But the book answer is a)the electron trajectory will be less curved than the proton trajectory. Please help!