A proton moves perpendicularly to a magnetic field that has a magnitude of 1.22×10−2

T.
The charge on a proton is 1.60×10−19 C.
What is the speed of the particle if the
magnitude of the magnetic force on it is 4.79×
10−14N?
Answer in units of m/s

To find the speed of the particle, we can use the equation for the magnetic force experienced by a charged particle:

F = qvB

Where:
F is the magnetic force (given as 4.79×10^(-14) N)
q is the charge on the particle (given as 1.60×10^(-19) C)
v is the speed of the particle (what we want to find)
B is the magnitude of the magnetic field (given as 1.22×10^(-2) T)

We can rearrange the equation to solve for v:

v = F / (qB)

Substituting the given values:

v = (4.79×10^(-14) N) / ((1.60×10^(-19) C) × (1.22×10^(-2) T))

v = 4.79×10^(-14) / (1.60×10^(-19) × 1.22×10^(-2)) m/s

v ≈ 2.476 × 10^5 m/s

Therefore, the speed of the particle is approximately 2.476 × 10^5 m/s.

To find the speed of the particle, we need to use the formula for the magnetic force on a charged particle in a magnetic field:

F = qvB

Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field strength

In this case, we're given:
F = 4.79 × 10^(-14) N (magnitude of the magnetic force)
q = 1.60 × 10^(-19) C (charge on a proton)
B = 1.22 × 10^(-2) T (magnitude of the magnetic field)

We can rearrange the formula to solve for v:

v = F / (qB)

Let's substitute the values:

v = (4.79 × 10^(-14) N) / (1.60 × 10^(-19) C × 1.22 × 10^(-2) T)

Now, let's multiply and divide the numbers:

v = (4.79 × 10^(-14) N) / (1.60 × 10^(-19) C × 1.22 × 10^(-2) T)
v = 4.79 × 10^(-14) / (1.60 × 10^(-19) × 1.22 × 10^(-2)) (N / C × T)

Now, let's simplify the expression:

v = (4.79 × 10^(-14)) / (1.60 × 1.22 × 10^(-19) × 10^(-2)) (N / C × T)
v = (4.79 × 10^(-14)) / (1.95 × 10^(-21)) (N / C × T)

And finally, let's calculate the value:

v ≈ 2.46 × 10^7 m/s

Therefore, the speed of the particle is approximately 2.46 × 10^7 m/s.

F=qvB =evB

v=F/eB=...