A helium nucleus (2 protons, 2 neutrons) is traveling at a velocity of 8.13 x 10^6 m/s a distance of 1.07 cm away from a wire carrying a current of 4.66 A. The helium nucleus is traveling in a direction parallel to the current in the wire. How much acceleration does the helium nucleus experience due to the magnetic force from the current in the wire?
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Q: See the attached figure for this problem. There is a charge of 6.24 x 10^-6 C moving at velocity…
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- A negatively charged ion with an excess electron moving with a velocity vx = 1.095 x103 m/s is introduced into a magnetic field Bz = 4 mTesla perpendicular to the ion’s direction of motion. The ion travels in a semi circle of diameter d = 8 cm before striking a detector. What is the ion?An electron experiences a magnetic force of magnitude 4.6 x 1015. N when moving at an angle of 60° with respect to a magnetic field of magnitude 3.5 x 103T. Find the speed of the electron. Select one: 9.49 x 10° m/s 1.19 x 10° m/s 5.25 x 10° m/s O22.1 x 10 m/s 2.40x10m/sA proton moving at 5.50 x 106 m/s through a magnetic field of magnitude 1.77 T experiences a magnetic force of magnitude 7.30 × 10-13 N. What is the angle between the proton's velocity and the field? Step 1 The magnitude of the force on a moving charge in a magnetic field is FB = qvB sin 8, and, solving for the angle, we have the following. 8 sin -1 = sin-¹ B qvB 1.60 x 10-19 c)([ x 10-13 N |× 106 m/s) (
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