A particle of charge 1.0 \times 10^{-16} {C} moves through a uniform magnetic field \overrightarrow{ {B}= {B}_{0}(\hat{ {i}+4 \hat{ {j}) {T}. The particle velocity at some instant is \overrightarrow{ {V}=(2 \hat{ {i}+4 \hat{ {j}) {m} {s}^{-1} and the magnetic force acting on it is 3 \times 10^{-16} \hat{ {k} {N}. The magnitude of {B}_{\circ} is

AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Physics

Question : 110 of 160

Marks: +1, -0

A particle of charge 1.0×10−16C moves through a uniform magnetic field

→

=B0(

∧

+4‌

∧

)‌T. The particle velocity at some instant is

→

=(2

∧

+4‌

∧

)‌m‌s−1 and the magnetic force acting on it is 3×10−16

∧

N. The magnitude of B∘ is

Go to Question:

Prev Question

Next Question