When does a nebula become a star?
1 answer:
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The complete question is;
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
