A) For the situation described in figure, the magnetic field is directed into th

Question

A) For the situation described in figure, the magnetic field is directed into the page ? and varies as B = d t3 ? a t2 + b , where d = 2 T/s 3 , a = 4.36 T/s 2 , and b = 0.69 T, and R = 1.7 cm, r = 4.19 cm.

Calculate the magnitude of the force exerted on an electron located at point P when t = 1.3 s. Answer in units of N

B) What is the direction of the force? 1. perpendicular to that of the electric field and magnetic field 2. same as that of the magnetic field 3. opposite to that of the magnetic field 4. opposite to that of the electric field 5. same as that of the electric field 6. Insufficient information is given.

C) At what (non-zero) time is this force equal to zero? Answer in units of 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 = d t3 a t2 + b , where d = 2 T/s 3 , a = 4.36 T/s 2 , and b = 0.69 T,

B = 2t3 - 4.36t2 + 0.69

dB = 6t2 - 8.72t, at time t=1.3s, so

dB = -1.196

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

E = -1.196*(pi*R^2) /(2pi*r1)

R = 1.7 cm = 1.7*10^-2 m

r1 = 4.19 cm = 4.19*10^-2 m

so,

E = -1.196*(pi*(1.7*10^-2 m)^2) /(2pi*4.19*10^-2 m)

E = -4.12*10-3

Now we can plug this back into the equation for Force,



Force = -4.12*10-3 *(1.602*10^-19 C) = 6.60*10-22

Using Right Hand Rule, we know this force is clockwise

Part C,
The force is equal to zero dB/dt = 0
dB = 6t2 - 8.72t = 0

Solving for t, we get t = 8.72/6 => 1.45 s

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