Two long parallel conductors $$S_1$$ and $$S_2$$ are separated by a distance $$10$$ cm and carrying currents of $$4$$ A and $$2$$ A respectively. The conductors are placed along $$x$$-axis in $$X-Y$$ plane. There is a point $$P$$ located between the conductors (as shown in figure).
A charge particle of $$3\pi$$ coulomb is passing through the point $$P$$ with velocity $$\vec{v} = (2\hat{i} + 3\hat{j})$$ m s$$^{-1}$$.
The force acting on the charge particle is $$4\pi \times 10^{-5}(-x\hat{i} + 2\hat{j})$$ N. The value of $$x$$ is

Step 1: Magnetic field at point P

$$B_{\text{net}}=B_1-B_2$$

$$B_1=\frac{\mu_0\cdot4}{2\pi(0.04)},\quad$$

$$B_2=\frac{\mu_0\cdot2}{2\pi(0.06)}$$

$$B_{\text{net}}=\frac{\mu_0}{2\pi}\left(\frac{4}{0.04}-\frac{2}{0.06}\right)$$

= $$\frac{\mu_0}{2\pi}\left(100-\frac{100}{3}\right)$$

=$$\frac{\mu_0}{2\pi}\cdot\frac{200}{3}$$

Direction: $$-\hat{k}$$

$$\vec{B} = \frac{\mu_0}{2\pi}\cdot \frac{200}{3}(-\hat{k})$$

Step 2: Magnetic force on charge

$$\vec{F} = q(\vec{v} \times \vec{B})$$

= $$3\pi \left[(2\hat{i} + 3\hat{j}) \times \left(\frac{\mu_0}{2\pi}\cdot \frac{200}{3}(-\hat{k})\right)\right]=3π[(2i^+3j^​)×(2πμ0​​⋅\ \frac{\ 200}{3}​(−k^))]$$

= $$3\pi \cdot \frac{\mu_0}{2\pi} \cdot \frac{200}{3} \left[2\hat{j} - 3\hat{i}\right]$$

Step 3: Simplify
Using $$\mu_0 = 4\pi \times 10^{-7}$$4:

$$\vec{F} = (4\pi \times 10^{-7})(100)\left(-3\hat{i} + 2\hat{j}\right)$$

= $$4\pi \times 10^{-5} \left(-3\hat{i} + 2\hat{j}\right)$$

Compare with given:

$$\vec{F} = 4\pi \times 10^{-5}(-x\hat{i} + 2\hat{j})$$

⇒ x = 3

Final Answer:

3