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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:
Answer:
x = 7.14 meters
Explanation:
It is given that,
Current in wire 1,
Current in wire 2,
Distance between parallel wires, r = 25 cm
Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :
Let the distance is x from wire 1. So,
x = 7.14 meters
So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.
Answer:
Explanation:
Given
Child falls a height of
Applying conservation of energy we can write
Loss in Potential energy=Gain in kinetic energy
Thus child reaches a velocity of 4.42 m/s