Proton move in a circle of radius 3.10 cm in a 0.566T magnetic field.What value of electric field could make their path straight
To determine the value of the electric field needed to make the path of a proton straight in a magnetic field, we can utilize the Lorentz force equation:
F = q(E + v x B)
Where,
F represents the net force on the particle,
q is the charge of the particle (which is the charge of a proton = 1.6 x 10^-19 C),
E is the electric field,
v is the velocity of the charged particle, and
B is the magnetic field.
In this case, we want to find the electric field that can nullify the force exerted by the magnetic field, resulting in a straight path. Therefore, the net force should be zero.
Since the proton is moving in a circle, the centripetal force acting on it is given by:
F_c = (mv^2) / r
Where,
m is the mass of the proton (approximately 1.67 x 10^-27 kg),
v is the velocity of the proton, and
r is the radius of the circular path.
By equating the centripetal force and the Lorentz force, we can find the value of the electric field:
F_c = F
(mv^2) / r = q(E + v x B)
Simplifying and rearranging the equation gives:
E = - (v x B)
Now we can substitute the values into the equation:
E = - (vB)
To solve for the magnitude of the electric field, we only need to calculate the magnitude of the cross product between the velocity and the magnetic field. Since the velocity and magnetic field are perpendicular to each other (as the magnetic field causes the circular motion), the cross product simplifies to the product of their magnitudes:
E = vB
Plugging in the given values:
E = (3.10 cm) x (0.566 T)
To ensure consistency in units, we need to convert the radius to meters:
E = (0.0310 m) x (0.566 T)
E ≈ 0.0175 V/m
Therefore, the value of the electric field required to make the proton's path straight is approximately 0.0175 V/m.