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A proton (q = 1.60 × 10−19 C, m = 1.67 × 10−27 kg) moves in a uniform magnetic field
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University Physics with Modern Physics (14th Edition)
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- A magnetic field directed into the page changes with time according to B = 0.030 0t2 + 1.40, where B is in teslas and t is in seconds. The field has a circular cross section of radius R = 2.50 cm (see Fig. P23.28). When t = 3.00 s and r2 = 0.020 0 m, what are (a) the magnitude and (b) the direction of the electric field at point P2?arrow_forwardA proton moving in the plane of the page has a kinetic energy of 6.00 MeV. A magnetic field of magnitude H = 1.00 T is directed into the page. The proton enters the magnetic field with its velocity vector at an angle = 45.0 to the linear boundary of' the field as shown in Figure P29.80. (a) Find x, the distance from the point of entry to where the proton will leave the field. (b) Determine . the angle between the boundary and the protons velocity vector as it leaves the field.arrow_forwardA particle moving downward at a speed of 6.0106 m/s enters a uniform magnetic field that is horizontal and directed from east to west. (a) If the particle is deflected initially to the north in a circular arc, is its charge positive or negative? (b) If B = 0.25 T and the charge-to-mass ratio (q/m) of the particle is 40107 C/kg. what is ±e radius at the path? (c) What is the speed of the particle after c has moved in the field for 1.0105s ? for 2.0s?arrow_forward
- A proton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle (charge +2e, mass 4mp) are accelerated from rest through a common potential difference V. Each of the particles enters a uniform magnetic field B, with its velocity in a direction perpendicular to B. The proton moves in a circular path of radius p. In terms of p, determine (a) the radius rd of the circular orbit for the deuteron and (b) the radius r for the alpha particle.arrow_forwardWhy is the following situation impossible? Figure P28.46 shows an experimental technique for altering the direction of travel for a charged particle. A particle of charge q = 1.00 C and mass m = 2.00 1015 kg enters the bottom of the region of uniform magnetic field at speed = 2.00 105 m/s, with a velocity vector perpendicular to the field lines. The magnetic force on the particle causes its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an angle from its original direction. The magnetic field has magnitude B = 0.400 T and is directed out of the page. The length h of the magnetic field region is 0.110 m. An experimenter performs the technique and measures the angle at which the particles exit the top of the field. She finds that the angles of deviation are exactly as predicted. Figure P28.46arrow_forwardA wire 2.80 m in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) 60.0, (b) 90.0, and (c) 120.arrow_forward
- One long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y = 0.280 m, z = 0). (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of 2.00 C is moving with a velocity of 150iMm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (c) What If? A uniform electric field is applied to allow this particle to pass through this region undetected. Calculate the required vector electric field.arrow_forwardWithin the green dashed circle shown in Figure P23.28, the magnetic field changes with time according to the expression B = 2.00t3 − 4.00t2 + 0.800, where B is in teslas, t is in seconds, and R = 2.50 cm. When t = 2.00 s, calculate (a) the magnitude and (b) the direction of the force exerted on an electron located at point P1, which is at a distance r1 = 5.00 cm from the center of the circular field region. (c) At what instant is this force equal to zero?arrow_forwardA spacecraft is in 4 circular orbit of radius equal to 3.0 104 km around a 2.0 1030 kg pulsar. The magnetic field of the pulsar at that radial distance is 1.0 102 T directed perpendicular to the velocity of the spacecraft. The spacecraft is 0.20 km long with a radius of 0.040 km and moves counter-clockwise in the xy-plane around the pulsar. (a) What is the speed of the spacecraft? (b) If the magnetic field points in the positive z-direction, is the emf induced from the back to the front of the spacecraft or from side to side? (c) Compute the induced emf. (d) Describe the hazards for astronauts inside any spacecraft moving in the vicinity of a pulsar.arrow_forward
- Two long coaxial copper tubes, each of length L, are connected to a battery of voltage V. The inner tube has inner radius o and outer radius b, and the outer tube has inner radius c and outer radius d. The tubes are then disconnected from the battery and rotated in the same direction at angular speed of radians per second about their common axis. Find the magnetic field (a) at a point inside the space enclosed by the inner tube r d. (Hint: Hunk of copper tubes as a capacitor and find the charge density based on the voltage applied, Q=VC, C=20LIn(c/b) .)arrow_forwardAn alpha-particle ( m=6.641027kg , q=3.21019C ) travels in a circular path of radius 25 cm in a uniform magnetic field of magnitude 1.5 T. (a) What is the speed of the particle? (b) What is the kinetic energy in electron-volts? (c) Through what potential difference must the particle be accelerated in order to give it this kinetic energy?arrow_forward
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