**Problem 4: Particle Motion in Electric and Magnetic Fields**Consider an electron at rest that is accelerated by an electric potential of 150 V. After this acceleration, the particle enters a region with a uniform magnetic field of 0.5 T, which is oriented perpendicular to the motion of the electron. *Question:* What force is experienced by the electron if its velocity is perpendicular to the magnetic field?**Explanation:**- *Electric Potential and Acceleration*: An electron accelerated through an electric potential difference gains kinetic energy. - *Magnetic Force*: When a charged particle moves through a magnetic field, it experiences a force if the velocity is perpendicular to the field. This force can be calculated using the formula: \[ F = qvB \sin(\theta) \] where: - \( F \) is the magnetic force, - \( q \) is the charge of the particle, - \( v \) is the speed of the particle, - \( B \) is the magnetic field strength, - \( \theta \) is the angle between the velocity and the magnetic field (90 degrees in this case, so \(\sin(90^\circ) = 1\)).

Answered: Image /qna-images/answer/55e6e7d2-0107-4d15-b4e6-4681bf43baec.jpg

Answered: 4) Consider an electro at rest and is.…

Similar questions

An electron moves with a velocity of 10^7 m/s in the x-y plane at an angle of 45∘ to both the +x and +y axes. There is a magnetic field of 2.0 T in the +y direction.1)Determine the magnitude of the magnetic force on the electron. (Express your answer using two significant figures.)
An proton moving with absolute velocity v = 4*10^6 m/s throgh a magnetic field of 3 T , a magnetic force act on it by FB = 6.7*10^-13 N. a) what is the angle θ between the velocity of the proton and the field. b) calculate the total force on the proton in case we applied an electric field by E=20 N/C
A proton is moving in a circular orbit of radius 28.9 cm with a speed of 3.7*106m/s in a uniform magnetic field perpendicular to the velocity of the proton. Find the magnetic field in which the proton is moving. mp = 1.67×10-27 kg, charge of proton = 1.61×10-19C. Also calculate the time period of the proton. Calculate the magnetic field and the time period to two decimal places. Submit the magnetic field as your answer. You are required to show your work in the file you submit.
An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 4.00 mT. The angular momentum of the electron about the center of the circle is 6.00 x 10-25 1.s. (a) Determine the radius of the circular path. cm (b) Determine the speed of the electron. m/s
Fields of an electron At a particular instant a proton is at the origin, moving with velocity m/s. At this instant: (a) What is the electric field at location m, due to the proton? E = N/C (b) What is the magnetic field at the same location due to the proton? B= T
23. A singly charged ion of mass m is accelerated from rest by a potential difference AV. It is then deflected by a uniform magnetic field (perpendicular to the ion's velocity) into a semicircle of radius R. Now a doubly charged ion of mass m' is accelerated through the same potential difference and deflected by the same magnetic field into a semicircle of radius R' = 2R. What is the ratio of the masses of the ions?
An electron is accelerated through 2500 V from rest and then enters a region where there is a uniform 1.60 T magnetic field. What are the maximum and minimum magnitudes of the magnetic force acting on this electron?