A proton, moving with velocity = -234 + 6662 m/s encounters a magnetic field B =...
A proton, moving with velocity v = -234 +6662 m/s encounters a magnetic field B = 3.252 UT. What is the radius of the circular part of the proton's trajectory?
Question 15 6.25 pts A proton, moving with velocity ✓ = –234+ 6662 m/s encounters a magnetic field B = 3.252 u T. What is the radius of the circular part of the proton's trajectory?
A proton, moving with velocity v=-235x+666z m/s encounters a magnetic field B=3.25z uT. What is the radius of the circular part of the proton's trajectory?
a proton is moving with speed of 5.0x10^18 m/s the proton encounter a magnetic field of 0.82T and whose direction makes an angle of 30.0 degrees with respect to the proton's velocity. determine the magnitude of the force on the proton Newt. assume the charge of a proton is 1.6x10^-19 C
A proton moving at 4.50 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.80 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest.) a. in ° b. in °
A proton moving at 7.00 x 106 m/s through a magnetic field of magnitude 2.6 T experiences magnetic force of magnitude 6.9x10-13 N. What is the angle between the proton's velocity and the field? 6.2° 76.3° 13.7° 44.9°
Question 13 3 pts A proton moving at 7x109 m/s through a magnetic field of magnitude 26 T experiences a magnetic force of magnitude 6.9x10- N What is the angle between the proton's velocity and the field? O 6.2 44.99 13.7 76.39 None of the above
Problem 3. a. An electron moving at 4.00x103 m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40*10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers. b. A proton moves at 7.50x107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength?
Question 1 2 pts (a) A. roton moving at a velocity of 2.00x106m/s encounters a magnetic field B = 1.20x10-5T whose direction is 45.0° from the direction of its velocity. What is the magnetic force on the proton? Answer: x10-18N (b) An electron is moving north in a region where a magnetic field points up. Which direction is the magnetic force on the electron (north, south, east, west, up or down)? Answer:
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proton having an initial velocity of 17.2i Mm/s enters a uniform magnetic field of magnitude 0.290 T with a direction perpendicular to the proton's velocity. It leaves the field-filled region with velocity -17.2] Mm/s (a) Determine the direction of the magnetic field (b) Determine the radius of curvature of the proton's path while in the field (c) Determine the distance the proton traveled in the field (d) Determine the time interval for which the proton is in...