A charge of 4.0 μ C is moving with a velocity of 4.0 × 10⁶ ms⁻¹ along the positive y axis under a magnetic field B⃗ of strength (2 k̂) T. The force acting on the charge is x î N. The value of x is __________.

Solution

Physics sir · 3 steps

Use ĵ × k̂ = î in qv⃗ × B⃗; then multiply the magnitudes.

1. Lorentz force. There is no electric field, so the only force on the moving charge is the magnetic force. Write the velocity and the field as vectors: v⃗ = 4.0 × 10⁶ ĵ m s⁻¹ and B⃗ = 2 k̂ T.

   F⃗ = q(v⃗ × B⃗) = (4.0 × 10⁻⁶)(4.0 × 10⁶)(2)(ĵ × k̂)

2. Cross product. The cyclic order î → ĵ → k̂ gives ĵ × k̂ = +î. The magnitudes multiply to 4.0 × 10⁻⁶ × 4.0 × 10⁶ × 2 = 16 × 2 = 32.
3. So. F⃗ = 32 î N, so x = 32.

The common slip: -32. Taking B⃗ × v⃗ instead of v⃗ × B⃗ flips the direction and gives -32. The order of the cross product matters.

The idea to keep. For magnetic force, keep the order v⃗ × B⃗ and use the cyclic rule î → ĵ → k̂: with the cycle gives a plus sign, against it gives a minus.

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