In a very long straight conductor, a current $I=\left(10^{3} \mathrm{~A} / \mathrm{s}\right) \cdot t$ dependent on time flows. At a distance of $l=5 \mathrm{~cm}$ from the straight conductor, a square-shaped conductor frame with side length $r=0.2 \mathrm{~m}$ and resistance $R=10^{-2} \Omega$ is placed. The straight conductor lies in the plane of the square. (We neglect the self-inductance of the frame!)

![](https://cdn.mathpix.com/cropped/2024_05_02_11b01584fe8b6b6986e1g-1.jpg?height=282&width=263&top_left_y=312&top_left_x=926)

a) How does the induced voltage in the square-shaped conductor depend on time?

b) When will the magnetic induction at the center of the square be zero?
a) The induced voltage $U$ in the frame can be calculated from the relationship

$$
U=\frac{\mathrm{d} \Phi}{\mathrm{d} t}
$$

where $\Phi$ is the magnetic flux induced by the straight conductor inside the frame.

![](https://cdn.mathpix.com/cropped/2024_05_02_d9f665420b39afbbcd66g-1.jpg?height=268&width=249&top_left_y=359&top_left_x=927)

It is known that at a distance $x$ from a very long straight conductor carrying a current $I$,

$$
\left|\mathbf{B}_{v}\right|=\frac{\mu_{0} I}{2 \pi x}
$$

( $\mu_{0}$ is the permeability of free space).

$$
\text { 1985-01-040-1.eps }
$$

To determine $\Phi$, we divide the frame into thin rectangles of width $(r / n)$ and height $r$ (see the figure). In the $i$-th rectangle, for sufficiently large $n$, $\mathbf{B}_{v}$ is approximately constant, and according to (2), for a suitable $\xi_{i}$,

$$
\left|\mathbf{B}_{v}\right|=\frac{\mu_{0} I}{2 \pi \xi_{i}}
$$

where $r_{i-1}<\xi_{i}<r_{i}, r_{i}=1+i r / n$.

Thus, the total flux through the frame for sufficiently large $n$ is approximated by

$$
\sum_{i=1}^{n} \frac{\mu_{0} I}{2 \pi \xi_{i}} \cdot \frac{r}{n} \cdot r=\frac{\mu_{0} I r}{2 \pi} \sum_{i=1}^{n} \frac{1}{\xi_{i}} \cdot \frac{r}{n}
$$

It is known that

$$
\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{\xi_{i}} \cdot \frac{r}{n}=\int_{l}^{l+r} \frac{1}{x} d x=\ln (l+r)-\ln l=\ln \left(1+\frac{r}{l}\right)
$$

Therefore,

$$
\Phi=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{\mu_{0} I}{2 \pi \xi_{i}} \cdot \frac{r}{n} r=\frac{\mu_{0} I r}{2 \pi} \ln \left(1+\frac{r}{l}\right)
$$

Thus, from (1) and (3),

$$
U=\frac{\mu_{0} r}{2 \pi} \ln \left(1+\frac{r}{l}\right) \frac{\mathrm{d} I}{\mathrm{~d} t}
$$

According to the problem, $I=k \cdot t\left(k=10^{3} \mathrm{~A} / \mathrm{s}\right)$, so from (4) with our data, the induced voltage in the frame is

$$
U=6.44 \cdot 10^{-5} \mathrm{~V}
$$

The voltage is constant over time, so the current $I_{k}$ flowing in the frame is also time-independent,

$$
I_{k}=\frac{U}{R}=\frac{6.44 \cdot 10^{-5} \mathrm{~V}}{10^{-3} \Omega}=6.44 \cdot 10^{-2} \mathrm{~A}
$$

b) According to Lenz's law, the magnetic induction $\mathbf{B}_{v}$ created by the straight conductor at the center $O$ of the frame and the magnetic induction $\mathbf{B}_{k}$ created by the current flowing in the frame are in opposite directions, so the resultant induction is

$$
\mathbf{B}=\mathbf{B}_{v}-\mathbf{B}_{k}=\mathbf{0}
$$

if

$$
\left|\mathbf{B}_{v}\right|=\left|\mathbf{B}_{k}\right| \text {. }
$$

The $\mathbf{B}_{k}$ induction can be calculated using the Biot-Savart law. Due to symmetry, it is sufficient to calculate the $\mathbf{B}_{k}^{A B}$ value created by the $A B$ segment of the frame, then

$$
\left|\mathbf{B}_{k}\right|=4 \cdot\left|\mathbf{B}_{k}^{A B}\right|=4 \cdot \frac{\mu_{0} I_{k}}{4 \pi} \int_{A B} \frac{\cos \alpha}{a^{2}} \mathrm{~d} y=\frac{2 \mu_{0} I_{k}}{\pi r} \int_{-\pi / 4}^{\pi / 4} \cos \alpha \mathrm{d} \alpha=2 \sqrt{2} \frac{\mu_{0} I_{k}}{\pi r}
$$

where we used the relationships $a=\frac{r}{2 \cdot \cos \alpha}, y=\frac{r}{2} \operatorname{tg} \alpha, \mathrm{d} y=\frac{r}{2 \cdot \cos ^{2} \alpha} \mathrm{d} \alpha$ (see the first figure).

The $\mathbf{B}_{v}$ induction at point $O$ is, similar to (2),

$$
\left|\mathbf{B}_{v}\right|=\frac{\mu_{0} I}{2 \pi(l+r / 2)}
$$

Substituting this and the (7) expression into (6) and using $I=k \cdot t$, we get:

$$
\left|\mathbf{B}_{v}\right|=\frac{\mu_{0} k \cdot t}{2 \pi(l+r / 2)}=2 \sqrt{2} \frac{\mu_{0} I_{k}}{\pi r}=\left|\mathbf{B}_{k}\right|
$$

Rearranging, the desired time is

$$
t=\frac{4 \cdot \sqrt{2}(l+r / 2)}{k \cdot r} I_{k}=\frac{4 \cdot \sqrt{2} \cdot 0.15 \mathrm{~m}}{10^{3}(\mathrm{~A} / \mathrm{s}) \cdot 0.2 \mathrm{~m}} \cdot 6.44 \cdot 10^{-2} \mathrm{~A}=2.73 \cdot 10^{-4} \mathrm{~s}
$$

after the time corresponding to $I=0$.