At a particular instant, a proton, far from all other objects, is located at the origin. The proton is traveling with velocity (−8 × 106, 0, 0)m/s. Consider the electric and magnetic fields at observation point (8 × 10−¹⁰,5 × 10−¹⁰,0)m caused by this proton. What is the electric field at the observation point? 1.37E9 9 1.37E9 X 9 0 <
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- A rocket zooms past the earth at v = 2.0 x 10 m/s. Scientists on the rocket have created the electric and magnetic fields shown in the figure. (Figure 1) Assume that B=0.60 T and E = 6.0x105 V/m. Figure B E 2.0 × 10 m/s 1 of 1 ▼ Part A What is the electric field strength measured by an earthbound scientist? Express your answer using two significant figures. VE ΑΣΦ Submit Part B Up Down O Left What is the direction of the electric field measured by an earthbound scientist? B = Request Answer Right Into the page Submit Out of the page Part C Stad Request Answer ? V/m What is the magnetic field strength measured by an earthbound scientist? Express your answer using two significant figures. VE ΑΣΦ LITER ? T1. A magnetic field has a magnitude of 1.20 x 10 T, and an electric field has a magnitude of 4.70 x 10° N/C. Both fields point in the same direction. A positive 2.5 µC charge moves at a speed of 3.00 x 10° m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge. N5.
- An electron encounters an E and B fields.B field is given by B = 0.1 j T.Electron experiences a force F = (9.6 x 10^-14 i – 9.6 x 10^-14 k)Find the electric field encountered by the electron. The velocity of the electron isv = 5 x 10^6 i m/s.can you please ans (a) (b) (c)?What is the magnitude and direction of the acceleration on the electron in a 3T magnetic field ( -k) and an 800 V/m (-j) electric field, shown below (neglecting gravity), if its velocity is 300 m/s? X X X B= 3.0 T X X X X X X X X X X E=800 V/m 3.512 EE+13 m/s2 (j) -3.512 EE+13 m/s2 (j) 3.512 EE-13 m/s2 (k) 2.489 EE+13 (j) -2.489 EE+13 (j)
- A proton is traveling through a region of space containing both a magnetic and electric field. At one instant in time, the velocity is U = (13500 Î + +38500) m/s and the acceleration is a = (4.39 k) x 10 m/s². If B=(9.21 k) mT, what is the magnitude of the electric field? i Save for Later N/C Submit AnswerA charged particle moves through a velocity selector at constant velocity. The velocity selector is configured with "crossed" electric and magnetic fields of magnitude E = 1.50 x 104 N/C and B = 0.4 T. %3D Hint a. What is the velocity of the charged particle? Velocity of the charged particle is x 104 m/s. b. When the electric field is turned off, the charged particle travels in a circular path of radius 6 mm, as it travels through the magnetic field (still at B = 0.4 T). What is the mass-to-charge ratio of the particle? Hint for (b) Mass-to-charge ratio of the part is kg/C. (Use the "E" notation to enter your answer in scientific notation. For example, to enter 3.14 × 10´ 12".) - 12 enter "3.14E-(unrealistic) E/M Forces 6. A particle of charge q = 0.25C and mass m = 1.0kg is moving with a velocity of v = (2,0,0)m/s. The (8,8,8)N/C and a magnetic field of B = (0,0, –2)T. charged particle finds itself in an electric field of E What will be the initial vector acceleration of the particle?