A charged particle carrying charge 1 μC is moving
with velocity ( 2î + 3ĵ + 4k̂ ) ms –1 . If an external
magnetic field of ( 5î + 3ĵ - 6k̂ )× 10 –3 T exists in the region where the particle is moving then the
force on the particle is F⃗ × 10 –9 N. The vector F⃗ is :

Solution

Physics sir · 4 steps

Do the cross product with plain numbers first, then gather all the powers of ten at the end.

1. Lorentz force. The magnetic force is F⃗ = q (v⃗ × B⃗). Here q = 10⁻⁶ C, and B⃗ carries a factor of 10⁻³ T. Keep both factors aside and cross the plain vectors (2, 3, 4) and (5, 3, -6).

   |î, ĵ, k̂; 2, 3, 4; 5, 3, -6| = î (-18-12) - ĵ (-12-20) + k̂ (6-15)

2. Cross product. This gives -30î + 32ĵ - 9k̂. Watch the middle term: -ĵ (-32) becomes +32ĵ.
3. Collect the powers. Multiply by the two factors you set aside: 10⁻⁶ × 10⁻³ = 10⁻⁹. So F⃗ = (-30î + 32ĵ - 9k̂) × 10⁻⁹ N.
4. So. The vector in the bracket is -30î + 32ĵ - 9k̂, which is option (c).

The idea to keep. For q v⃗ × B⃗, cross the bare numbers first and multiply the powers of ten separately. The minus sign on the ĵ term is where most errors happen.

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