Lenz's Law — Question 1

Question 1

The induced current need not reverse when the applied field crosses zero

A fixed loop of area A and resistance R has normal +z out of the page. Its uniform applied field is $B_z=B_0(1-t/T)$, with B0 and T positive. Neglect self-inductance.

1. Find the current sense before and after t=T.

2. For A=.0200 square metre, B0=.300 T, T=.100 s and R=6.00 ohms, calculate its value.

Question 1 - Solution

The induced current need not reverse when the applied field crosses zero

(a) Method and calculation

$$\dot\Phi=A\dot B_z=-AB_0/T<0.$$$$\mathcal E=-\dot\Phi=AB_0/T>0\Rightarrow\boxed{\text{counterclockwise for both }t<T\text{ and }t>T}.$$

(b) Interpretation and further calculation

$$i=\frac{AB_0}{RT}=\frac{(.0200)(.300)}{(6.00)(.100)}.$$$$\boxed{i=.0100\,\mathrm A}.$$

Physical check

The induced field opposes the change, not necessarily the instantaneous applied field.