Protons move in a circle of radius 6.10 cm in a 0.566-T magnetic field. What value of electric field could make their paths straight?
To make the paths of protons straight, we need to find the electric field that cancels out the magnetic force acting on them. The magnetic force experienced by a charged particle moving in a magnetic field is given by the equation:
F = qvB sin(θ)
Where:
F is the magnetic force
q is the charge of the particle (in this case, the charge of a proton is +1.6 x 10^-19 C)
v is the velocity of the particle
B is the magnetic field strength
θ is the angle between the velocity vector and the magnetic field vector
In this case, the protons are moving in a circle. The magnetic force is always perpendicular to the velocity vector, so the angle θ is 90 degrees (or π/2 radians).
Now, let's find the velocity of the proton. The velocity of a particle moving in a circle can be calculated using the equation:
v = ωr
Where:
v is the velocity
ω is the angular velocity
r is the radius of the circle
The angular velocity ω can be calculated using the equation:
ω = v / r
Rearranging this equation, we get:
v = ωr
Substituting the value of ω into the equation for v, we get:
v = (v / r) * r
v = v
Therefore, the velocity of the proton is equal to the velocity of the proton.
Now, let's substitute these values into the equation for the magnetic force:
F = qvB sin(θ)
F = (1.6 x 10^-19 C) * v * (0.566 T) * sin(π/2)
Since sin(π/2) = 1, the equation becomes:
F = (1.6 x 10^-19 C) * v * (0.566 T)
To cancel out this magnetic force, an electric force of equal magnitude but opposite direction is required. The electric force on a charged particle moving in an electric field is given by the equation:
F = qE
Where:
F is the electric force
q is the charge of the particle (in this case, the charge of a proton is +1.6 x 10^-19 C)
E is the electric field strength
To make the paths of the protons straight, the electric force must cancel out the magnetic force, so:
F_electric = F_magnetic
qE = (1.6 x 10^-19 C) * v * (0.566 T)
Now, let's rearrange this equation to solve for E:
E = [(1.6 x 10^-19 C) * v * (0.566 T)] / q
Substituting the value of the proton velocity (v = v) into the equation, we get:
E = [(1.6 x 10^-19 C) * v * (0.566 T)] / q
Finally, substituting the values of the charge of a proton (q = 1.6 x 10^-19 C), the magnetic field strength (B = 0.566 T) and the radius of the circle (r = 6.10 cm = 0.0610 m), we can calculate the value of the electric field (E):
E = [(1.6 x 10^-19 C) * v * (0.566 T)] / (1.6 x 10^-19 C)
Simplifying the equation, we find:
E = v * (0.566 T)
Therefore, the value of the electric field required to make the paths of the protons straight is equal to the product of the proton velocity and the magnetic field strength:
E = v * (0.566 T)