At a particular instant in the,in a region of space where E=0 and B=3i Wbm^-1 , a 2 kg particle of charge 1 C moves with a velocity of 2ims^-1.What is the particle's acceleration due to magnetic field?
To find the particle's acceleration due to the magnetic field, we can use the equation:
F = q(v x B)
where:
F = force experienced by the particle
q = charge of the particle
v = velocity of the particle
B = magnetic field
First, let's calculate the force experienced by the particle:
F = q(v x B)
= (1 C) * (2 m/s) * (3i T)
= 6i C m/s T
Since the force is given as F = ma, where m is the mass of the particle and a is the acceleration, we can calculate the particle's acceleration:
F = ma
6i C m/s T = (2 kg) * a
Now, to find the acceleration, we need to convert the force from newtons to newton-meter per second squared (N m/s^2):
1 T = 1 N/(A m)
1 C = 1 A s
So, we have:
6i C m/s T = [(2 kg) * a] * (1 N/(A m)) * (1 A s)
Simplifying,
6i C m/s T = (2 kg) * a * (1 N m/(A s))
Now, let's isolate the acceleration (a) by dividing both sides by (2 kg) * (1 N m/(A s)):
6i C m/s T / [(2 kg) * (1 N m/(A s))] = a
Finally, simplifying the units and the expression:
[6i C m/s T / (2 kg N m/(A s))] = a
[6i C m^2/(kg s)] = a
Therefore, the particle's acceleration due to the magnetic field is 6i C m^2/(kg s).
Note: In this calculation, I have assumed that the particle is moving perpendicular to the magnetic field and that the force experienced by the particle is solely due to the magnetic field.