Respuesta :
Answer:
a = 603.59 m/s^2
Explanation:
from the data given . the rate of change in magnetic field is as follow
[tex]\frac{dB}{dt} = 280 G/s = 280 \times 10^{-4} T/s[/tex]
from the faraday's law of induction , the expression for the induced emf in region of radius r as follow
[tex]\epsilon = \frac{d \phi}{dt}[/tex]
[tex]\int E.dl = \frac{d(BA)}{dt}[/tex]
[tex]E(2\pi r)= \pi r^2 \frac{dB}{dt}[/tex]
[tex]E = \frac{r}{2} \frac{dB}{dt}[/tex]
electric field at point P_1 as follow
[tex]E = \frac{r}{2} \frac{dB}{dt}[/tex]
[tex]E = \frac{1.5\times 10^{-2}}{2} 280 \times 10^{-4}[/tex]
[tex]E = 6.3\times 10^{-6} V/m[/tex]
from newton 2nd law of motion, the acceleration of proton is
F = ma
qE = ma
[tex]a = \frac{qE}{m}[/tex]
[tex]a = \frac{1.6 \times 10^{-19} (6.3\times 10^{-6})}{1.67\times 10^{-27}}[/tex]
a = 603.59 m/s^2
The acceleration of the object depends on the velocity and time. The acceleration of the proton at point P1 is [tex]20.12\times 10^{-3} \;\rm m/s^2[/tex].
What is acceleration?
The acceleration can be defined as the rate of change in the motion of the object with respect to time.
Given that radius of the cylindrical region is 6.5 cm and the rate of change in the magnetic field is as follow,
[tex]\dfrac {dB}{dt} = 280 \;\rm G/s[/tex]
The emf produced in the cylindrical region is given as below.
[tex]\int Edl = \dfrac {d(BA)}{dt}[/tex]
Where E is the electric field and A is the area of the region.
[tex]E (2\pi r) = \pi r^2 \dfrac {dB}{dt}[/tex]
[tex]E = \dfrac {r}{2} \dfrac{dB}{dt}[/tex]
At point P1 = 1.5 cm from the center, the electric field is given as,
[tex]E = \dfrac {1.5 \times 10^{-2}}{2} \times 280 \times 10^{-4}[/tex]
[tex]E = 2.1 \times 10^{-4}\;\rm V/m[/tex]
The force on the proton is given as below.
[tex]F = ma[/tex]
Where m is the mass and a is the acceleration of the proton.
The force on the proton is given by the electric field is,
[tex]F = qE[/tex]
Where E is the electric field at point P1 and q is the charge on the proton. We can write the force as,
[tex]ma = qE[/tex]
[tex]a = \dfrac {qE}{m}[/tex]
[tex]a = \dfrac {1.6\times 10^{-19} \times 2.1 \times 10^{-4}}{1.67 \times 10^{-27}}[/tex]
[tex]a = 20.12 \times 10^{-3}\;\rm m/s^2[/tex]
Hence we can conclude that the acceleration of the proton at point P1 is [tex]20.12\times 10^{-3} \;\rm m/s^2[/tex].
To know more about the acceleration, follow the link given below.
https://brainly.com/question/2437624.
