a) When t = 2.00 s, calculate the magnitude of the force exerted on an electron

Question

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 = 4.90 cm from the center of the circular field region.

(b) When t = 2.00 s, calculate the direction of the force exerted on an electron located at point P1, which is at a distance r1 = 4.90 cm from the center of the circular field region.

Tangent to the electric field line passing through point P1 and clockwise.Tangent to the electric field line passing through point P1 and counterclockwise.    The magnitude is zero.

(c) At what instant is this force equal to zero? (Consider the time after t = 0 s.)

Explanation / Answer

Force = E*q   (*where E=electric field, q=charge (of electron) = 1.602*10^-19 C )

To find force, first find E.

Emf = integral( E*ds) = - dB/dt

dB/dt = B*A (*where A=area)

A = (pi*R^2)

So, we can now solve for E, by taking the derivative of Emf:

integral( E*ds) = B*A becomes...

E = dB*A/ds

**note ds = 2pi*r1

We find dB by taking the derivative of B:

B = 9.00t3 1.00t2 + 0.800,

dB = 27t2 - 2t, at time t=2, so

dB = 104

Plugging dB and A into the equation for E we get:

E = 104*(pi*R^2) /(2pi*r1)

R = 2.45 cm = 2.45*10^-2 m

r1 = 4.9 cm = 4.9*10^-2 m

so, E = 104*(pi*(2.45*10^-2 m)^2) /(2pi*4.9*10^-2 m) = 0.637

Force = 0.637*(1.602*10^-19 C) = 1.02*10^-19 N

b) Using Right Hand Rule, we know this force is clockwise

c) The force is equal to zero dB/dt = 0

dB = 27t2 - 2t = 0

Solving for t, we get t = 2/27 = 0.074 s

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