Answer:
The acceleration is [tex]a = 3.45*10^{3} m/s^2[/tex]
Explanation:
From the question we are told that
The radius is [tex]d = 6.5 cm = \frac{6.5}{100} = 0.065 m[/tex]
The magnitude of the magnetic field is [tex]B = 5.5 T[/tex]
The rate at which it decreases is [tex]\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s[/tex]
The distance from the center of field is [tex]r = 1.5 cm = \frac{1.5}{100} = 0.015m[/tex]
According to Faraday's law
[tex]\epsilon = - \frac{d \o}{dt}[/tex]
and [tex]\epsilon = \int\limits {E} \, dl[/tex]
Where the magnetic flux [tex]\o = B* A[/tex]
E is the electric field
dl is a unit length
So
[tex]\int\limits {E} \, dl = - \frac{d}{dt} (B*A)[/tex]
[tex]{E} l = - \frac{d}{dt} (B*A)[/tex]
Now [tex]l[/tex] is the circumference of the circular loop formed by the magnetic field and it mathematically represented as [tex]l = 2\pi r[/tex]
A is the area of the circular loop formed by the magnetic field and it mathematically represented as [tex]A= \pi r^2[/tex]
So
[tex]{E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}[/tex]
[tex]E = \frac{r}{2} [ - \frac{db}{dt} ][/tex]
Substituting values
[tex]E = \frac{0.015}{2} (24*10^{-4})[/tex]
[tex]E = 3.6*10^{-5} V/m[/tex]
The negative signify the negative which is counterclockwise
The force acting on the proton is mathematically represented as
[tex]F_p = ma[/tex]
Also [tex]F_p = q E[/tex]
So
[tex]ma = qE[/tex]
Where m is the mass of the the proton which has a value of [tex]m = 1.67 *10^{-27} kg[/tex]
[tex]q = 1.602 *10^{-19} C[/tex]
So
[tex]a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}[/tex]
[tex]a = 3.45*10^{3} m/s^2[/tex]
