An +9.1 C charge moving at 0.71 m/s makes an angle of 45∘ with a uniform, 1.9 T magnetic field. What is the magnitude of the magnetic force ? that the charge experiences?
Magnetic force on a moving charge is given by the relation:
F = qvBsin(the)
= 9.1*0.71*1.9*sin45
= 8.68 N (answer)
An +9.1 C charge moving at 0.71 m/s makes an angle of 45∘ with a uniform,...
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A proton moving in the positive x direction with a speed of 9.1 105 m/s experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 1.8 10-13 N that points in the positive z direction. Determine the magnitude and direction of the magnetic field. magnitude T
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Find the magnitude of the force on a -5.58-C charge moving with velocity 1.89 m/s north in a uniform magnetic field of 3.59 T east. Answer in units of N.
An electron moving at 4.5x106 m/s through a magnetic field and experiences a magnetic force of magnitude 8.4x10-13 N. If the angle between the proton's velocity and the field is 27°, find the magnitude magnetic field (in T)? (Charge and mass of an electron are 1.6x10-19 C, and 9.1x10-31 kg) 2.57 3.27 1.31 0.94
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If a 0.001 C charge is has velocity of 10 m/s at an angle of 30° relative to a magnetic field of strength 20 T, then what is the magnitude of the magnetic force? IN 0.1N 10N 20N
A proton moving in a uniform magnetic field with v⃗ 1=1.13×106ı^m/s experiences force F⃗ 1=1.70×10−16k^N. A second proton with v⃗ 2=2.06×106ȷ^m/s experiences F⃗ 2=−4.17×10−16k^N in the same field. What is the magnitude of B⃗ ? What is the direction of B⃗ ? Give your answer as an angle measured ccw from the +x-axis.