Question

10. If any particle having charge 10 C moves along
north-east direction with velocity 12root2 m/s at a
particular instant it enters in uniform magnetic field
of 10 T (vertically upward). The force on this charge
is
(1) 1200JEN (North-east)
(2) 1200 N (North-east)
(3) 1200/2N (South-east)
(4) 1200 N (South-east)​

Answer 1

Given info : If any particle having charge 10 C moves along  north-east direction with velocity 12√2 m/s at a  particular time it enters in uniform magnetic field  of 10 T (vertically upward).

To find : the force on this charge is ...

solution : velocity of the charge, v = 12√2( cos45° i + sin45° j) = 12i + 12j

magnetic field, B = 10k

now magnetic force on the charge, F = q ( v × B)

= 10C {(12i + 12j) × 10k}

= 100 {-12j + 12i}

= 1200i - 1200j

now the magnitude of magnetic force, |F| = 1200√(1² + 1²) = 1200√2N

therefore the force on the charge is 1200√2N south east direction.

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