Two long parallel conductors S₁ and S₂ are separated by a distance 10 cm and carrying currents of 4A and 2A 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 3π coulomb is passing through the point P with velocity v⃗ = (2î + 3ĵ) m/s; where î and ĵ represents unit vector along x y axis respectively.
The force acting on the charge particle is 4π × 10⁻⁵( - xî + 2ĵ) N. The value of x is :

Solution

Physics sir · 4 steps

Find the net B at P from both wires first, then let qv⃗ × B⃗ do the rest.

1. Field from S1. P is 4 cm below S₁, and the current in S₁ flows along +x. By the right-hand rule, î × (-ĵ) = -k̂, so the field points along -z. Its size is B₁ = (μ₀ I)/(2π d) = (2 × 10⁻⁷ × 4)/0.04 = 2 × 10⁻⁵ T.
2. Field from S2. P is 6 cm above S₂, and î × ĵ = +k̂, so this field points along +z. Its size is B₂ = (2 × 10⁻⁷ × 2)/0.06 = 2/3 × 10⁻⁵ T. The two fields oppose each other, so the net field is B⃗ = -4/3 × 10⁻⁵ k̂ T.

   F⃗ = q v⃗ × B⃗ = 3π (2î + 3ĵ) × (-4/3 × 10⁻⁵ k̂)

3. Cross product. Use î × k̂ = -ĵ and ĵ × k̂ = î. This gives (2î + 3ĵ) × k̂ = 3î - 2ĵ. So F⃗ = 3π × 4/3 × 10⁻⁵ (-3î + 2ĵ) = 4π × 10⁻⁵ (-3î + 2ĵ) N.
4. So. Compare this with 4π × 10⁻⁵(-xî + 2ĵ). The ĵ part matches, and x = 3, which is option (c).

The idea to keep. Between two parallel wires carrying current the same way, their fields point opposite ways, so you subtract them. Fix the direction of B first, then the cross product gives the force.

Answer from the NTA final answer key. Solution by PrepWiser.