When a charged particle moves at an angle of 17.0o with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 2F?
To solve this problem, we can use the relationship between the magnetic field, velocity of the charged particle, the angle between them, and the magnetic force experienced by the particle.
The magnetic force on a charged particle moving in a magnetic field can be determined using the formula:
F = q * v * B * sin(θ)
where:
- F is the magnetic force on the particle
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnitude of the magnetic field
- θ is the angle between the velocity vector and the magnetic field vector
According to the problem, when the charged particle moves at an angle of 17.0° with respect to the magnetic field, it experiences a magnetic force of magnitude F. Let's call this angle θ₁.
Now, we need to find the angle θ₂ at which the particle will experience a magnetic force of magnitude 2F, while moving at the same speed. Let's substitute the given information into the formula above:
F = q * v * B * sin(θ₁)
We can rearrange the formula to solve for the magnitude of the magnetic field:
B = F / (q * v * sin(θ₁))
Now, let's substitute this expression for B into the formula and solve for θ₂:
2F = q * v * (F / (q * v * sin(θ₁))) * sin(θ₂)
Dividing both sides of the equation by q * v, we get:
2 = F / (q * v * sin(θ₁)) * sin(θ₂)
Multiplying both sides of the equation by q * v * sin(θ₁), we have:
2 * q * v * sin(θ₁) = F * sin(θ₂)
Dividing both sides of the equation by F, we get:
2 * q * v * sin(θ₁) / F = sin(θ₂)
Finally, taking the inverse sine (or arcsine) of both sides of the equation, we get:
θ₂ = arcsin(2 * q * v * sin(θ₁) / F)