In order to go through a velocity selector undisturbed, the magnetic and electric force on charge q=7.1 mC need to cancel out (as in, they have to point in opposite directions and their magnitudes have to be the same). Determine the velocity of the charge that can travel through a velocity selector that provides electric field of magnitude E=2.99x10³ V/m and magnetic field B=7.71 mT. Provide your answer in km/s.
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- Asapkg. Consider the mass spectrometer shown schematically in the figure below. The electric field between the plates of the velocity selector is 945 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.900 T. Calculate the radius r of the path for a singly charged ion with mass m = 2.06 x 10-26 mm X x P x r Bo, in X x X X X X Photographic plate Bin Velocity selector x x x x + + 1₂ E x x x x X--- --- V X X x || 9 XA I x I x X X x x x X XCrossed electric and magnetic fields are established over a certain region. The magnetic field is 0.590 T vertically downward. The electric field is 3.48 × 106 V/m horizontally east. An electron, traveling horizontally northward, experiences zero net force from these fields and so continues moving in a straight line. What is the electron’s speed?
- q10An electromagnetic flowmeter applies a magnetic field of 0.20 T to blood flowing through a coronary artery at a speed of 15 cm/s. What force is felt by a chlorine ion with a single negative charge?A horizontal magnetic field of 1 x 10-11 T (Teslas) exists at an angle of 30 degrees to the direction of the current in a straight, horizontal wire 75cm long. If the wire carries a current of 15A (amps): a. What is the magnitude of force on the wire, expressed in Newtons (N)? b. What angle would be required for the force to be half the value calculated in part (a), assuming nothing else is changed?
- Protons move in a circle of radius 5.80 cm in a 0.55T magnetic field. What value of electric ficld could make their paths straight? Mass of a proton = 1.67 x 10"kg. Charge of a proton = 1.6 x 10"C.A circular region 19.0 m in radius is filled with an electric field perpendicular to the face of the circle. The magnitude of the field in the circle varies with radius and time as E(r,t) = Arcos(at) where A 1000. V/m2 and -5.00 x 10's. What is the maximum value of the magnetic field at the edge of the region? TAn electron moves in a straight line at speed 250 m/s along the positive y-axis through uniform electric and magnetic fields. The magnetic field is directed into the page and has magnitude 2 T. Neglect gravity. The mass of the electron is 9.1 x 10³1 kg. a. What is the electric field (specify both magnitude and direction)? b. Now imagine the electric field has been turned off. What will be the radius of the electron's subsequent path?
- An electron is accelerated by a potential difference of 3000V and enters a region of a uniform magnetic field. As a result the electron bends along a path with a radius of curvature of 26.0 cm. a-Find the speed of the electron in [Mm/s]as it enters the magnetic field. b-Find the speed of the electron as it leaves the magnetic field in [Mm/s].A charged particle moves through a velocity selector at constant velocity. The velocity selector is configured with "crossed" electric and magnetic fields of magnitude E = 2.00 × 10¹ N/C and B = 1 T. Hint a. What is the velocity of the charged particle? Velocity of the charged particle is 2.00 x 10¹ 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= 1 T). What is the mass-to-charge ratio of the particle? Hint for (b) Mass-to-charge ratio of the particle is 2.5E-8 x kg/C. (Use the "E" notation to enter your answer in scientific notation. For example, to enter 3.14 x 10-¹2, enter "3.14E-12".)Please help. I don't get this.