Concept explainers
(a)
The magnetic field at origin.
(a)
Answer to Problem 13P
The magnetic field at origin is
Explanation of Solution
Given:
The charge is
The velocity is
Formula used:
The expression for magnetic field is given by,
Calculation:
The magnetic field at origin is calculated as,
Conclusion:
Therefore, the magnetic field at origin is
(b)
The magnetic field at
(b)
Answer to Problem 13P
The magnetic field is
Explanation of Solution
Calculation:
The magnetic field is calculated as,
Conclusion:
Therefore, the magnetic field is
(c)
The magnetic field at
(c)
Answer to Problem 13P
The magnetic field is
Explanation of Solution
Calculation:
The magnetic field is calculated as,
Conclusion:
Therefore, the magnetic field is
(d)
The magnetic field at
(d)
Answer to Problem 13P
The magnetic field is
Explanation of Solution
Calculation:
The magnetic field is calculated as,
Conclusion:
Therefore, the magnetic field is
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Chapter 27 Solutions
Physics For Scientists And Engineers Student Solutions Manual, Vol. 1
- 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_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_forwardSolenoid A has length L and N turns, solenoid B has length 2L and N turns, and solenoid C has length L/2 and 2N turns. If each solenoid carries the same current, rank the magnitudes of the magnetic fields in the centers of the solenoids from largest to smallest.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_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_forwardDoes increasing the magnitude of a uniform magnetic field through which a charge is traveling necessarily mean increasing the magnetic force on the charge? Does changing the direction of the field necessarily mean a change in the force on the charge?arrow_forward
- A 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_forwardA long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of magnitude 4.0105 T that is directed vertically downward, what is tire resultant magnitude of the magnetic field 20 cm above the wire? 20 cm below the wire?arrow_forwardA mass spectrometer (Fig. 30.40, page 956) operates with a uniform magnetic field of 20.0 mT and an electric field of 4.00 103 V/m in the velocity selector. What is the radius of the semicircular path of a doubly ionized alpha particle (ma = 6.64 1027 kg)?arrow_forward
- Consider an electron rotating in a circular orbit of radius r. Show that the magnitudes of the magnetic dipole moment and the angular momentum L of the electron are related by: = L=e2marrow_forwardA circular coil of radius 5.0 cm is wound with five turns and carries a current of 5.0 A. If the coil is placed in a uniform magnetic field of strength 5.0 T, what is the maximum torque on it?arrow_forwardA long, straight wire lies on a horizontal table and carries a current of 1.20 μA. In a vacuum, a proton moves parallel to the wire (opposite the current) with a constant speed of 2.30 × 104 m/s at a distance d above the wire. Ignoring the magnetic field due to the Earth, determine the value of d.arrow_forward
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