Answered: A proton (with mass m, = 1.67 x 1027… | bartleby

Related questions

Question

A proton (with mass m, = 1.67 x 1027 kg, charge q, = 1.6 x 10-19 C) moving with a
speed of 0.80 x 105 m/s, enters the Earth's magnetic field (B= 55.0 µT, Northbound). If
the proton is shot eastward, the magnetic force acting on it should be
O 7.04 x 1019 N West-bound
7.04 x 1019 N upward
8.08 x 1016 N downward
7.04 x 10 19 N downward

Expert Solution

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

A charge q = 4 µC moves with velocity v = 30 i km/s in a magnetic field B = (0.2 j + 0.5 k) T. Find the magnetic force on the charge. (24 j + 60 k) mN (−24 j + 60 k) mN 65 i mN (−60 j + 24 k) mN (60 j − 24 k) mN
An electron moves at a velocity of 2.20 x 10^6 m/s in a direction perpendicular to a magnetic field of strength 1.89 T. How much force will the electron experience?
A beam of electrons moving with speed v = 2.6 × 10’m/s enter and then leave the rectangular region of the magnetic field, as shown below. The trajectory of electrons is a quarter of a circle of radius R= 0.020 m. Mass of electron m = 9.11 x 10-31 kg, charge of electron charge q = 1.6 × 10-19 C). The magnitude and direction of the magnetic field are: %3D %3D the region of the magnetic field a. 0.45 T, perpendicular to the page, into the page Ob. 0.74 T, perpendicular to the page, out of the page Oc. 0.45 T, perpendicular to the page, out of the page С. O d. 0.074 T, perpendicular to the page, into the page e. 0.68 T, perpendicular to he page, into the page
A particle of charge 18 µC feels a force of 2.4 × 10−4 N when it moves at a speed of 24 m/s in a direction perpendicular to a magnetic field. (a) What is the strength of the magnetic field in this region? (b) What is the force on the particle when it moves with a speed of 6.3 m/s at an angle of 25o relative to the direction of the magnetic field?
This time I1 = 5.03, I2 = 1.44 A, and the two wires are separated by 9.70 cm. Now consider the charge q = 7.27 x 10^-6 C, located a distance of 4.27 cm to the right of wire I2, moving to the right at speed v = 24.6 m/s. What is the magnitude of the total magnetic force on this charge? 1.19E-09 N 9.52E-10 N 2.49E-09 N 1.99E-09 N
A proton is moving along in the +y-direction with a speed of 5.2 × 105 m/s experiences a magnetic force of 8.5 × 10−4 N out of the page. A second proton moves along in the −x-direction at the same speed but experiences no magnetic force. What is the strength and direction of the magnetic field?[Answer: B⃗ = −1.0 × 1010 T xˆ]
A proton (charge q = 1.6 × 10-19 C) moving with velocity 3 = (-5.0 × 10°i)m/s enters a region in which the magnetic field is B = 0.30k T. The magnitude and the direction of the magnetic force acting on the %3D %3D %3D proton are: Hint: 1.0 pN = 10-12 N %3D a. 24 pN, in thenegative y direction b. 24 pN, in the positive y direction O c. 1.5 pN, in the positive x direction С. d. 24 pN, in the negative z direction e. 15 pN, in the positive y direction
Current I = 30 A flows in a straight wire. An ion with charge Q = 3.2e-19 C is moving toward the wire, at distant r = 0.10 m, with speed v = The magnetic force on the ion is antiparallel to the current. What is the magnitude of the force? (in N) 6.00×106 m/s. A: 3.682x10-17 B: 4.897x10-17 OC: 6.513x10-17 OD: 8.662x10-17 E: 1.152x10-16 OF: 1.532x10-16 OG: 2.038×10-16 OH: 2.710×10-16
A proton (with mass mp = 1.67 x 10-27 kg, charge qp = 1.6 x 10-19 C) moving with a speed of 0.80 x 105 m/s, enters the Earth’s magnetic field (B= 55.0 μμT, Northbound). If the proton is shot eastward, the magnetic force acting on it should be Group of answer choices 8.08 x 10-16 N downward 7.04 x 10-19 N West-bound 7.04 x 10-19 N upward 7.04 x 10-19 N downward