A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

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A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B⃗ =(1.63T)i^+(0.980T)j^?

Physics

2 answers:

ValentinkaMS [3.4K] · Answer 1 · 2026-03-09T04:05:00+03:00

The particle’s acceleration magnitude and direction in a uniform magnetic field $B =(1.63T)i$ ^ $+(0.980T)j$ ^ is $\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}$

Explanation:

A particle weighing 1.81×10−3 kg with a charge of 1.22×10−8 C has a velocity v⃗ =(3.00×104m/s)j^ at a specific moment. What are the acceleration's magnitude and direction from a uniform magnetic field $B =(1.63T)i$ ^ $+(0.980T)j$ ^?

A charged particle possesses an electric charge. Electric charge is a property that results in force attraction or repulsion in an electromagnetic field. A uniform magnetic field occurs when the magnetic field lines are straight and parallel, ensuring a consistent magnetic force throughout the area.

According to Newton's second law, the force can be defined by:

$F=ma$

The magnetic force is given by

$F= qv \times B$

$ma = qv \times B$

$a = \frac{qv \times B}{m}$

By substituting the earlier values, we find:

$a = \frac{(1.22 \times 10^{-8} C) (3 \times 10^{4} m/s) (1.63 T ) (\hat{j} \times \hat{i})}{1.81 \times 10^{-3} kg} = -(0.330 m/s^2) \bold{\hat{{k}}}$

Thus, the particle's resulting acceleration in a uniform magnetic field $B =(1.63T)i$ ^ $+(0.980T)j$ ^ is confirmed to be $\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}$

Learn further about the particle's acceleration