Exploring electromagnetism 1!

First we calculate the magnetic field due to the infinite conductor in a perpendicular distance $l$ from it: $\vec{B}=-\dfrac{\mu I}{2 \pi l} \hat{k}$

Then we calculate the emf in the conductor $EF$ : $\begin{aligned} \epsilon&=\int_{a}^{b} (\vec{v} \times \vec{B})\cdot \mathrm{d}\vec{l}\\ &=\dfrac{v \mu I}{2 \pi} \int_{a}^{b} \dfrac{\mathrm{d} l}{l}\\ &=\dfrac{v \mu I}{2 \pi} \ln\left(\dfrac{b}{a}\right) \end{aligned}$

Next, we calculate the induced current: $\begin{aligned} I'&=\dfrac{\epsilon}{R}\\ &=\dfrac{v \mu I}{2 \pi R} \ln\left(\dfrac{b}{a}\right) \end{aligned}$

Then we calculate the magnetic force: $\begin{aligned} \vec{F_M}&=I'\int_{a}^{b} \mathrm{d}\vec{l} \times \vec{B}\\ &=\dfrac{I' \mu I}{2\pi} \int_{a}^{b} \dfrac{\mathrm{d} l}{l} \hat{j}\\ &=\dfrac{I' \mu I}{2\pi} \ln\left(\dfrac{b}{a}\right) \hat{j} \\ &=\dfrac{v}{R} \left(\dfrac{\mu I}{2 \pi} \ln\left(\dfrac{b}{a}\right)\right)^2 \hat{j} \end{aligned}$

Finally, we muts apply another force $\vec{F}$ with the same magnitude tham $\vec{F_M}$ but in the opposite direction ( $-\hat{j}$ ) to keep moving the conductor with the constant velocity. Substituting values we get: $F=\dfrac{10\text{ m/s}}{10\;\Omega} \left(\dfrac{4\pi \times 10^{-7} \frac{\text{N}}{\text{A}^2} \times 10\text{ A}}{2 \pi} \ln\left(\dfrac{20\text{ m}}{2\text{ m}}\right)\right)^2 \\ F \approx 2.12075 \times 10^{-11}\text{N}$

So, our answer is $\boxed{21.2075}$ .