A proton moves perpendicular to a uniform magnetic field B at a speed of 2.10 x 107 m/s and experiences an acceleration of 1.60 x 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field.
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- An alpha particle has a positive charge that is double the charge of a proton, and a mass of 6.64 ✕ 10−27 kg. At a particular instant, it is moving with a speed of 5.00 ✕ 106 m/s through a magnetic field. At this instant, its velocity makes an angle of 62° with the direction of the magnetic field at the particle's location. The magnitude of the field is 0.230 T. (a) What is the magnitude of the magnetic force (in N) on the particle? (b) What is the magnitude of the particle's acceleration (in m/s2) at this instant?A proton moves perpendicularly to a uniform magnetic field at 3.0 ✕ 107 m/s and exhibits an acceleration of 1.0 ✕ 1013 m/s2 in the +x-direction when its velocity is in the +z-direction. Determine the magnitude and direction of the field.Needs Complete typed solution with 100 % accuracy.
- The solar wind is a thin, hot gas given off by the sun. Charged particles in this gas enter the magnetic field of the earth and can experience a magnetic force. Suppose a charged particle traveling with a speed of 7.26 x 106 m/s encounters the earth's magnetic field at an altitude where the field has a magnitude of 1.75 x 10-7 T. Assuming that the particle's velocity is perpendicular to the magnetic field, find the radius of the circular path on which the particle would move if it were (a) an electron and (b) a proton.In each of the figures below a charge q₁ = 11.3 nC is moving with a speed v = 2.0 * 107. The direction is pointing in the (a) right, (b) up, (c) left and (d) downward direction on a cartesian coordinate system as shown below. When the charge passes by the location (5.00cm, 5.00cm) what is the magnetic field vector (in component form) at the location (5.00cm, 0.00cm) for each situation? (a) (b) (c) P 91 Р V (d) 91 91Your answer is partially correct. One long wire lies along an x axis and carries a current of 57 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 7.2 m, 0), and carries a current of 45 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 2.0 m, 0)? Number i 0.000005956 Units PT
- Determine the magnitude of the force on an electron traveling 5.85×105 m/s horizontally to the east in a vertically upward magnetic field of strength 0.20 T . Express your answer to two significant figures and include the appropriate units. Determine the direction of the force on an electron.A uniform magnetic field B = 1.7 x 10-4 T exists in +x direction. A proton (q = +e, and mp = 1.67 x 10-27 kg) is shot through the field in the +y direction with a speed of 3.5 x 105 m/s. Find the magnitude and direction of the magnetic force FB on the proton.A proton moves perpendicular to a uniform magnetic field at a speed of 2.30 107 m/s and experiences an acceleration of 2.70 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.