An electron (with mass m, = 9.1x 10 31 kg, charge |qel = 1.6 x 10-19 C) moving with a speed of 2.50 x 105 m/s, enters the Earth's magnetic field (B= 55.0 µT, Northbound). If the electron moves westward, the magnetic force (FB) acting on it should be 2.2 x 10 18 N eastward 7.04 x 10 16 N upward 2.2 x 1018 N downward O 2.2 x 1018 N upward
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- A particle with a charge of q = -5.20 nC is moving in a uniform magnetic field of B =(-1.24 T ) 2. The magnetic force on the particle is measured to be F =(-7.60x10-7 N )âŷ.1. An electron has a velocity v = (2 x 106 m/s) i +(3.0x 106 m/s)j moves through a uniform magnetic field B = (0.030 T) ? - (0.15 T)j. (a) Find the force on the electron due to the magnetic field. (b) mass of electron is 9.11 x 10-31 kg what will be the radius of the circular path for the electron in the magnetic field?An electron is moving in a circular orbit of radius 2 mm in a uniform 0.50 T magnetic field perpendicular to the velocity of the electron. What is the speed of the electron? Consider: The charge of an electron = 1.6 x 10-19 C The mass of an electron = 9.1 x 10-31 kg !3! 878
- A particle with a charge of 2μC is moving at 3 x 106m/s perpendicularly through a magnetic field with a strength of 0.05T. What is the magnitude of the force on the particle?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 NA 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]
- ements valuación An electron moves in a uniform circular motion under the action of an external magnetic field perpendicular to the circular path. Consider that the charge and mass of the electron are respectively q = 1.6 x 10-¹⁹ C, m = 9.11 10-31 Kg, the velocity of the electron is v = = 2.8 × 107 and the magnitude of the external magnetic field is B = 2.1 x 10-³T. Calculate the radius R of the circle formed by the electron in its path during its displacement and choose the correct option. 5.5 cm O 7.5 cm 9.5 cm O 3.5 cm 1 pts Question 3Part 2) A proton moving with a speed 3.84 × 10° m/s through a 2.99 T magnetic field experiences a force of magnitude 1.59 x 10-12 N. (Mass of the proton = 1.673 × 10¬27 kg, charge 1.602 × 10-19 C). What is the angle between the proton's velocity and the magnetic field? Part 3) What is the proton's acceleration? a = m/s?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
- An electron (with mass mp = 9.1x 10-31 kg, charge ∣∣qe∣∣ = 1.6 x 10-19 C) moving with a speed of 2.50 x 105 m/s, enters the Earth’s magnetic field (B= 55.0 μμT, Northbound). If the electron moves westward, the magnetic force (FB) acting on it should be Group of answer choices 2.2 x 10-18 N eastward 2.2 x 10-18 N downward 7.04 x 10-16 N upward 2.2 x 10-18 N upwardA 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