A loop of wire lays flat on a horizontal surface in an area with uniform magnetic field directed vertically down. Magnetic field
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Answer:
As the capacitor is discharging, the current is increasing
Explanation:
Lets take
C= Capacitance
L=Inductance
V=Voltage
I= Current
The total energy E given as
We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL² will increases.
It means that when charge on the capacitor decreases then the current will increase.
As the capacitor is discharging, the current is increasing
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: