Question

a magnetic field, b, is given as a function of the distance, r, from the center of the wire as follows: (b = \frac{r}{r_0}b_0) when (rleq r_0), and (b=\frac{r_0}{r}b_0) when (r > r_0). here (r_0) and (b_0) are positive constants. sketch a graph of b against r, and use it to answer the following questions. as you do so, think about what the meaning of the constant (b_0) is. a. is b a continuous function of r? (enter yes or no. be sure that you can explain your answer.) b. is b a differentiable function of r? (enter yes or no. be sure that you can explain your answer.)

Explanation:

Step1: Check continuity at $r = r_0$

Left - hand limit as $r
ightarrow r_0^{-}$: $\lim_{r
ightarrow r_0^{-}}B=\lim_{r
ightarrow r_0^{-}}\frac{r}{r_0}B_0=\frac{r_0}{r_0}B_0 = B_0$. Right - hand limit as $r
ightarrow r_0^{+}$: $\lim_{r
ightarrow r_0^{+}}B=\lim_{r
ightarrow r_0^{+}}\frac{r_0}{r}B_0=\frac{r_0}{r_0}B_0 = B_0$. And $B(r_0)=\frac{r_0}{r_0}B_0 = B_0$. Since the left - hand limit, right - hand limit and the function value at $r = r_0$ are equal, $B$ is continuous at $r = r_0$ and for all $r>0$. So $B$ is a continuous function of $r$.

Step2: Check differentiability at $r = r_0$

Derivative for $r r_0$: $B'=-\frac{r_0B_0}{r^{2}}$. Left - hand derivative at $r = r_0$ is $\frac{B_0}{r_0}$ and right - hand derivative at $r = r_0$ is $-\frac{r_0B_0}{r_0^{2}}=-\frac{B_0}{r_0}$. Since the left - hand derivative and right - hand derivative at $r = r_0$ are not equal, $B$ is not a differentiable function of $r$.

Answer:

A. yes
B. no