College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- A nucleus with 40 protons travels at a velocity of 2.74 x 10^6 m/s through a region of uniform magnetic field strength 2.93 T. The nucleus experiences a magnetic force of 3.44E-11 N. What angle does the nucleus s velocity vector make with respect to the magnetic field direction? 42.0 degrees 21.0 degrees 37.8 degrees 54.6 degreesarrow_forwardPart (a) Calculate the numerical value of F in newtons.arrow_forwardTo see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 300-loop-per-meter circular coil 0.660 m in radius to create a 1.20-T field (typical of an MRI instrument) at its center with no iron present. The magnetic field of a proton is approximately like that of a circular current loop 0.650×10−15 m in radius carrying 2.1×104 A . What is the field at the center of such a loop? 2.39 x 103 A and 1.02 x 1013 T 2.39 x 10 3 A and 2.04 x 10 13 T 3.183 x 10 3 A and 2.04 x 10 13 T 3.183 x 10 3 A and 1.02 x 10 13 T None of the abovearrow_forward
- helparrow_forwardCalculate the radius of curvature of the path of an electron having a velocity of 6 × 10^7m/s. Perpendicular to a magnetic field of strength ? = 0.5 T.??????? = 9.11 × 10^−31kg???? = 1.6 × 10^−19carrow_forwardAn alpha particle (He++) is passing horizontally over your paper from right to left at 7.73 x 10^6 m/s when a large, uniform magnetic field of 12.44 T out of the paper is immediately turned on. The ion is caught in a circular path. An accurate and detailed sketch of the ion, magnetic field, and path of the ion is recommendedarrow_forward
- 1 Singly-charged lithium ions, liberated from a heated anode are accelerated by a difference of 400 V between anode and cathode. They then pass through a hole in the cathode into a uniform magnetic field perpendicular to their direction of motion. The magnetic flux density is 8 × 10-2 Wb/m² and the radii of the paths of the ions are 8.83 and 9.54 cm, respectively. Calculate the mass numbers of the lithium isotopes.arrow_forwardcan you please ans (a) (b) (c)?arrow_forward8. A charged particle moves in a horizontal plane with a speed of 8 70 106 . × m / s. When this particleencounters a uniform magnetic field in the vertical direction it begins to move on a circular path of radius15.9 cm. (a) If the magnitude of the magnetic field is 1.21 T, what is the charge-to-mass ratio (q/m) ofthis particle?arrow_forward
- Problem 5: An electron (q= -e = -1.60 × 10-19 C and me = 9.11 × 10-31 kg) in a cathode-ray tube is accelerated through a potential difference of 10 kV. Thus, the electron leaves the cathode-ray tube with a speed v = 5.93 × 107 m/sec. The electron then passes through the 2.0 cm wide region of uniform magnetic field (shown in the Figure above on the right). What field strength will deflect the electron by 10°? Answer: B = 0.0029 T. OV 10 kV 2.0 cm 10°arrow_forwardA small rod magnet has a magnetic field at a distance from it as large as Hom B = 2nz3 Where z is the distance from the magnet on the axis connecting the north pole and south, m is the moment of magnetic dipole. Assuming electrons orbit protons like small rod magnets, determine the m magnetic dipole of an electron orbiting an atom.arrow_forwardAn oxygen nucleus (8 protons, 8 neutrons) moves at a velocity of 6.55 x 10^6 m/s in a direction perpendicular to a uniform magnetic field. The oxygen nucleus is observed to move in a circular path of radius 3.45 cm. What is the strength of the magnetic field? 5.54 B 3.96 B 1.58 B 3.56 Barrow_forward
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