Question

A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^v→=(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^B→=(1.63T)i^+(0.980T)j^?

Answer 1

Answer:

-(0.330m/s² ) kˆ

Explanation:

given data:

Mass of particle 'm'= 1.81 x $10^{-3}$ kg

Velocity 'v'= (3.00 x$10^{4}$ m/s)j

Charge of particle 'q'= 1.22 x $10^{-8}$ C

Uniform magnetic field 'B' = (1.63iˆ + 0.980jˆ )T

In order to calculate particle's acceleration, we'll use Newton's second law of motion i.e F=ma

Also,the force a magnetic field exerts on a charge q moving with velocity v is called the magnetic Lorentz force. It is given by:

F = qv × B

F= ma = qV x B

a= $\frac{q(v*B)}{m}$ --->eq(1)

Lets determine the value of (v x B) first

v x B= (3.00 x$10^{4}$ m/s)j x (1.63iˆ + 0.980jˆ )

v x B= 4.89 x $10^{4}$

Plugging all the required values in eq(1)

a= [1.22 x $10^{-8}$ x (4.89 x $10^{4}$kˆ)] / 1.81 x $10^{-3}$

a=  -(0.330m/s² ) kˆ

-ve sign is representing the opposite direction