A proton is accelerated from rest through a 5.70-V potential difference.
(a) What is the proton's speed after it has been accelerated?
(b) What is the maximum magnetic field that this proton produces at a point that is 4.00 m from the proton? (Enter the magnitude.)
To find the answers to these questions, we can use the principles of electric potential energy and the relationship between electric and magnetic fields.
(a) The electric potential difference is given as 5.70 V. We know that the work done by the electric field on the proton is equal to its change in potential energy. Since the proton starts from rest, its initial kinetic energy is zero. Therefore, the electric potential energy gained by the proton is equal to its final kinetic energy.
The equation for electric potential energy is given by:
ΔPE = qΔV
Where ΔPE is the change in potential energy, q is the charge of the proton, and ΔV is the potential difference across which the proton is accelerated.
The kinetic energy of the proton is given by:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass of the proton, and v is its velocity.
Setting the change in potential energy equal to the final kinetic energy, we have:
qΔV = (1/2)mv^2
Solving for v, we get:
v = √(2qΔV / m)
To find the value of v, you'll need the charge of a proton (1.6 x 10^-19 C) and its mass (1.67 x 10^-27 kg).
(b) To find the maximum magnetic field produced by the proton at a point that is 4.00 m away, we can use the formula for the magnetic field produced by a moving charged particle:
B = (μ₀ / 4π) * (qv / r²)
Where B is the magnetic field, μ₀ is the permeability of free space (4π x 10^-7 T·m/A), q is the charge of the proton, v is its velocity, and r is the distance from the proton.
Substituting the values given in the question, we can calculate B using the known values for q, v (from part (a)), and r.
Finally, to get the magnitude of the magnetic field, take the absolute value of B.
Remember to use proper units throughout the calculations.