Chapter 40: Problem 16

Question: A long cylindrical conducting rod with radius R is centered on the x- axis as shown in Fig. 12 (chap 40). A narrow saw cut is made in the rod at x=b. A conduction current i increasing with time and given by $i~=~\alpha~t$ flows toward the right of the rod, $~\alpha$ is a positive constant of proportionality. At t=0 there is no charge on the cut faces near x=b.

1.

Find the magnitude of the charge on these faces as a function of time.

2.

Use Eq.I in table 2 to find E in the gap as a function of time.

3.

Sketch the field lines of B for r<R where r is the distance from the x- axis.

4.

Use Eqn IV in Table 2 to find B(r) in the gap for r<R

5.

Compare the above answer with B(r) in the rod for r<R

Solution: If  q (t)  is the charge on each face of the gap as a function of time, then

$$q (t) = q (0) + \int dt I(t) = {1\over 2} \alpha t^2 \ \ \ .$$ (20)

The charge density on the plates is thus

$$\sigma~(t)~=~{q(t)\over \pi R^2} \ \ \ ,$$ (21)

giving an electric field in the gap of

$$\vert{\bf E}\vert(t)~= ~{\sigma(t)\over\epsilon_0} = {\alpha t^2\over 2\pi\epsilon_0 R^2} \ \ \ .$$ (22)

From this we can use Amperes law to find the magnetic field,

$$\int {\bf B}\cdot d{\bf l}~=~\mu_0\epsilon_0~{d\Phi_E\over dt} \ \ \ .$$ (23)

$$\vert{\bf B}\vert 2\pi r~=~\mu_0\epsilon_0 \pi r^2~\vert{\bf E}\vert(t)~=~{\mu_0 \alpha t r \over 2 \pi R^2} \ \ \ .$$ (24)

Now, to compute B inside the rod, we assume a uniform current density, and use

$$\int~{\bf B}\cdot d{\bf l}~=~\mu_0 \int~{\bf j}\cdot d{\bf A}$$

$$\vert{\bf B}\vert 2\pi r = \mu_0 {I\over \pi R^2} \pi r^2$$

$$\vert{\bf B}\vert = { \mu_0 I r\over 2 \pi R^2 } \ =\ { \mu_0 \alpha t r\over 2 \pi R^2 } \ \ \ .$$ (25)

This is the same expression.