(a) When t = 2.00 s, calculate the magnitude of the force exerted on an electron located at point P1, which is at a distance r1 = 5.70 cm from the center of the circular field region.

Within the green dashed circle shown in the figure below, the magnetic field changes with time according to the expression B = 6.00t³ − 1.00t² + 0.800, where B is in teslas, t is in seconds, and R = 2.20 cm.

(a) When t = 2.00 s, calculate the magnitude of the force exerted on an electron located at point P1, which is at a distance r1 = 5.7 cm from the center of the circular field region.

I have attached the figure the question talks about.

Answer:

Force = 4.62 x 10^(-20) N

Explanation:

First of all,

∮E.ds = (d/dt)∮B.dA

E∮ds = (d/dt)B∮dA

dA is the area where B is not equal to zero.

Thus,

E•2πr = (d/dt)B•πR²

|E| = [(d/dt)B•πR²]/2πr

F = |qE| = q[(d/dt)B•πR²]/2πr

Where q is charge on electron = 1.6 x 10^(-19) C

F is magnitude of force exerted on electron

F = 1.6 x 10^(-19)[(d/dt)B•R²]/2r

F = 0.8 x 10^(-19)[(d/dt)B•R²]/r

Now, dB/dt = 18t² - 2t

Thus,

F = 0.8 x 10^(-19)[(18t² - 2t)•R²]/r

Thus, at t=2 and R=2.2cm = 0.022m and r = 5.7cm = 0.057m

Thus,

F = 0.8 x 10^(-19)[(18x2²) - (2x2)•(0.022²)]/(0.057)

F = (0.8 x 10^(-19) x 0.032912)/0.057 = 4.62 x 10^(-20) N

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-29T14:12:17+02:00

The magnitude of the force exerted on an electron located at point P1 is mathematically given as

F= 4.62 x 10^{-20} N

What is the magnitude of the force exerted on an electron located at point P1?

Question Parameter(s):

t = 2.00 s,

distance r1 = 5.70 cm

Generally, the equation for the Electric field is mathematically given as

[tex]E*2\pir = (d/dt)B*\pi R^2[/tex]

Therefore

[tex]F = |qE| \\F= q[(d/dt)B*\piR^2]/2\pi r[/tex]

F = 1.6 x 10^(-19)[(d/dt)B*R^2]/2r

F = 0.8 x 10^(-19)[(18t^2 - 2t)*R^2] /r

F = 0.8 x 10^(-19)[(18x2^2) - (2x2)*(0.022^2)]/(0.057)

F = (0.8 x 10^(-19) x 0.032912)/ 0.057

F= 4.62 x 10^{-20} N

In conclusion,the magnitude of the force is

F= 4.62 x 10^{-20} N