Protons move in a circle of radius 4.10 cm in a 0.570 T magnetic field. What value of electric field could make their paths straight?
Set Eq= Bqv
To find the value of the electric field that would make the proton's path straight, we need to use the equation that relates the electric field, magnetic field, charge, and velocity of a charged particle moving in a magnetic field.
The equation is given by:
F = q(E + v x B)
Where:
F is the net force on the particle,
q is the charge of the particle,
E is the electric field,
v is the velocity of the particle, and
B is the magnetic field.
In this case, since we want to make the proton's path straight, we need the net force on the proton to be zero. Therefore, we can set the equation equal to zero:
0 = q(E + v x B)
Since we are given the radius of the circle the proton is moving in and the magnetic field, we can determine the velocity of the proton using the formula for the centripetal force:
F_c = |q|vB = mv^2/r
Where:
F_c is the centripetal force,
m is the mass of the proton,
v is the velocity of the proton,
B is the magnetic field, and
r is the radius of the circle.
Rearranging this equation, we can solve for v:
v = (F_c * r) / (|q| * B)
Now, we know that the proton's velocity is perpendicular to the magnetic field, so v x B = vB. Substituting this into the earlier equation, we have:
0 = q(E + vB)
Since we know the value of q and the velocity, we can solve for the electric field:
E = -vB
Substituting the expression for v that we found earlier:
E = -[(F_c * r) / (|q| * B)] * B
Finally, we can plug in the values given in the problem and solve for the electric field:
E = -[(|q|vB * r) / (|q| * B)] * B
Simplifying:
E = -v * r
Now we can substitute the given values:
E = -[(2.0 x 10^-19 C) * (v) * (4.10 x 10^-2 m)] / [(1.6 x 10^-19 C)] * (0.570 T)
Calculating this expression will give us the value of the electric field that can make the proton's path straight.