Answer:
The acceleration of the proton is 9.353 x 10⁸ m/s²
Explanation:
Given;
speed of the proton, u = 6.5 m/s
magnetic field strength, B = 1.5 T
The force of the proton is given by;
F = ma = qvB(sin90°)
ma = qvB
where;
m is mass of the proton, = 1.67 x 10⁻²⁷ kg
charge of the proton, q = 1.602 x 10⁻¹⁹ C
The acceleration of the proton is given by;
Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²
The speed of a electron that is accelerated from rest through an electric potential difference of 120 V is
We know, that the energy of the system is always conserved.
Using the Law of Conservation of energy,
U=0
Here, K is the kinetic energy and U is the potential energy.
Now, substituting the formula of U and K, we get:
=0------(1)
Here,
m is the mass of the electron
v is the speed of the electron
q is the charge on the electron
V is the potential difference
Let and
represent the final and initial speed.
Here, =0
Solving for , we get:
=6.49m/s
To learn more about the conservation of energy, refer to:
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