Un electrón en un tubo de rayos catódicos acelera desde el reposo con una aceleración constante de 5.33x10¹²m/s² durante 0.150μs
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Explanation:
It is given that,
Magnetic field, B = 0.1 T
Acceleration,
Charge on electron,
Mass of electron,
(a) The force acting on the electron when it is accelerated is, F = ma
The force acting on the electron when it is in magnetic field,
Here,
So,
Where
v is the velocity of the electron
B is the magnetic field
v = 341250 m/s
or
So, the speed of the electron is
(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.
1.
For the first part, we just need to write the equation of the forces along two perpendicular directions.
We have actually only two forces acting on the car, if we want it to go around the track without friction:
- The weight of the car, mg, downward
- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)
By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:
(1)
(2)
where in the second equation, the term represents the centripetal force, with v being the speed of the car and R the radius of the track.
Dividing eq.(2) by eq.(1), we get the following expression:
2.
In this second situation, the cars moves around the track at a speed
This means that the centripetal force term
is now larger than before, and therefore, the horizontal component of the normal reaction, , is no longer enough to keep the car in circular motion.
This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes
And re-arranging for F,
(3)
But from eq.(2) in the previous part we know that
So, susbtituting into eq.(3),