A magnetic field directed into the page changes with time according to B = 0.020

Question

A magnetic field directed into the page changes with time according to B = 0.0200t2 + 1.40, where B is in teslas and t is in seconds. The field has a circular cross section of radius R = 2.50 cm (see figure below).

(a) When t = 3.85 s and r2 = 0.020 0 m, what is the magnitude of the electric field at point P2?
(b) When t = 3.85 s and r2 = 0.020 0 m, what is the direction of the electric field at point P2?

We use the general form of Faraday's law, and consider a circular integration path of radius r2. Due to symmetry, Faraday's law becomes

E(2piR2)= -d/dt (BA) = -A*(dB/dt)

Solving for the magnitude of the electric field, we have

|E|=A/(2piR2) d/dt (____________ t^2 + _____________)

    =(piR2^2)/(2piR2) (__________ t)

     = r2/2 (__________ t)

Explanation / Answer

B*2*pi*r2 = -A*dB/dt

dB/dt = 2*0.02*t + 0

B*2*pi*r2 = pi*r2^2*(2*0.02t )

B = r2*0.02*t

B = 0.02*0.02*3.85 = 0.00154 V/m

counter clock wise

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.