A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
1 answer:
A) d. 10T
When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by
where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.
This force acts as a centripetal force, keeping the particle in a circular motion - so we can write
which can be rewritten as
The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):
So, we get:
We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.
B)
The frequency of revolution of a particle in uniform circular motion is
where
f is the frequency
T is the period
We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:
T' = 10 T
Then its frequency of revolution will be:
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a) 22.5 m/s
The initial vertical velocity is given by:
where
u = 35 m/s is the initial speed
is the angle of projection of the ball
Substituting into the equation, we find
b) 26.8 m/s
The initial horizontal velocity is given by:
where
u = 35 m/s is the initial speed
is the angle of projection of the ball
Substituting into the equation, we find
c) 2.30 s
The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation
where
is the vertical velocity at time t
is the acceleration of gravity (negative because it is downward)
At the maximum height, the vertical velocity becomes zero, ; substituting, we find the time t at which this happens:
d) 25.8 m
The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:
where
s is the vertical displacement at time t
Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:
e) 123.3 m
In order to find how far does the ball lands, we have to consider the horizontal motion.
First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:
Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:
Therefore, the horizontal distance travelled during the whole motion is
So, the ball lands 123.3 m far from the initial point.