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An electron is shot into one end of a solenoid. As it enters the uniform magnetic field within the solenoid, its speed is 800 m/s and its velocity vector makes an angle of 30° with the central axis of the solenoid. The solenoid carries 4.0 A and has 8000 turns along its length. How many revolutions does the electron make along its helical path within the solenoid by the time it emerges from the solenoid’s opposite end? (In a real solenoid, where the field is not uniform at the two ends, the number of revolutions would be slightly less than the answer here.)
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Chapter 29 Solutions
Fundamentals of Physics, Volume 1, Chapter 1-20
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- How many turns must be wound on a flat, circular coil of radius 20 cm in order to produce a magnetic field of magnitude 4.0105 T at the center of the coil when the current through it is 0.85 A?arrow_forwardConsider a solenoid that is very long compared with its radius. Of the following choices, what is the most effective way to increase the magnetic field in the interior of the solenoid? (a) double its length, keeping the number of turns per unit length constant (b) reduce its radius by half, keeping the number of turns per unit length constant (c) overwrap the entire solenoid with an additional layer of current-carrying wirearrow_forwardThe accompanying figure shows a cross-section of a long, hollow, cylindrical conductor of inner radius r1= 3.0 cm and outer radius r2= 5.0 cm. A 50-A current distributed uniformly over the cross-section flows into the page. Calculate the magnetic field at r = 2.0 cm. r = 4.0 cm. and r = 6.0 cm.arrow_forward
- Assume the region to the right of a certain plane contains a uniform magnetic field of magnitude 1.00 mT and the field is zero in the region to the left of the plane as shown in Figure P22.71. An electron, originally traveling perpendicular to the boundary plane, passes into the region of the field. (a) Determine the time interval required for the electron to leave the field-filled region, noting that the electrons path is a semicircle. (b) Assuming the maximum depth of penetration into the field is 2.00 cm, find the kinetic energy of the electron.arrow_forwardA 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_forwardWithin the green dashed circle show in Figure P30.21, 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 P, which is at a distance r = 5.00 cm from the center of the circular field region. (c) At what instant is this force equal to zero? Figure P30.21arrow_forward
- An election moves through a uniform electric field E = (2.50i + 5.00j) V/m and a uniform magnetic field B = 0.400k T. Determine the acceleration of the electron when it has a velocity v = 10.0i m/s.arrow_forwardCalculate the magnitude of the magnetic field at a point 25.0 cm from a long, thin conductor carrying a current of 2.00 A.arrow_forwardA long, solid, cylindrical conductor of radius 3.0 cm carries a current of 50 A distributed uniformly over its cross-section. Plot the magnetic field as a function of the radial distance r from the center of the conductor.arrow_forward
- A proton moving horizontally enters a region where a uniform magnetic field is directed perpendicular to the proton’s velocity as shown in Figure OQ22.4. After the proton enters the field, does it (a) deflect downward, with its speed remaining constant; (b) deflect upward, moving in a semicircular path with constant speed, and exit the field moving to the left; (c) continue to move in the horizontal direction with constant velocity; (d) move in a circular orbit and become trapped by the field; or (e) deflect out of the plane of the paper? Figure OQ22.4arrow_forwardTwo 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_forwardWhy is the following situation impossible? A conducting rectangular loop of mass M = 0.100 kg, resistance R = 1.00 , and dimensions w = 50.0 cm by = 90.0 cm is held with its lower edge just above a region with a uniform magnetic field of magnitude B = 1.00 T as shown in Figure P30.34. The loop is released from rest. Just as the top edge of the loop reaches the region containing the field, the loop moves with a speed 4.00 m/s. Figure P30.34arrow_forward
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