1. Nov 13, 2014

### reed2100

1. The problem statement, all variables and given/known data
A circular loop of wire with radius 0.0300m and resistance 0.380Ω is in a region of spatially uniform magnetic field, as shown in the following figure(Figure 1). The magnetic field is directed into the plane of the figure. At t = 0, B = 0. The magnetic field then begins increasing, with B(t)=( 0.350T/s3)t3.

2. Relevant equations
ε = -N*(dΦB / dt)

3. The attempt at a solution
Area = pi*r^2 = pi(.03^2) = .0028 m^2

Magnetic field is perpendicular to plane of loop.

ΦB = (B→)⋅dA→ = BA cos(Θ)

B(t) is given at an instant where it equals 1.42 T, so solving 1.42 = .35 t^3 for t: (1.42/.35)^(1/3) = 1.59 seconds

I then reason that since t and be started at 0, at this given instant dB = 1.42 Tesla and dt = 1.59 seconds.

Since only the magnetic field changes, I can say dΦB/dt = (dB/dt)Acos(Θ)

Since Θ = 0 due to magnetic field being perpendicular to plane of loop, it simplifies to dΦB/dt = (dB/dt)A

So ε = 1*(dB/dt)A = (1.42/1.59)*.0028 = .0025 volts

However the question asks for current I in the loop, so I said ε = IR: thus ε/R = I by ohm's law

.0025 volts / .380 ohms = .0065 Amps.

This is incorrect, the correct answer was .0199 Amps and I'm not sure what I did wrong. Was I supposed to differentiate B(t) or something? I got the direction of the current in the loop correct, I understand Lemz law and recall the common right hand rule conventions. Any and all help is appreciated.

2. Nov 13, 2014

### TSny

Hello.
Dividing 1.42 T by 1.59 s gives the average rate of change of the B field during the time interval between t = 0 and t = 1.59 s. But you need the instantaneous rate of change of B at the instant when t = 1.59 s.

3. Nov 13, 2014

### reed2100

I see, so I would differentiate B(t) to get 3(.350 T)t^2, and then plug in t = 1.59 seconds to find the instantaneous rate of change of B at that instant, correct?

dB/dt is just saying the derivative of B with respect to t, so my derivation above and then plug in t =1.59 to find dB/dt at that instant of 1.59 seconds? That would be the change in magnetic field at that instant which when combined with it's propagation across the area .0028 m^2, is what induces the electromotive force in the loop.

So dB/dt with that said is 3(1.59^2)(.350) = 2.65

2.65(area of .0028 m^2) = .00742 volts

.00742 volts / .380 ohms = .0195 amps, and the answer was .019 amps for sig figs. Thank you so much! I see how I went wrong now.