(a)
The magnitude of magnetic field at center of loop.
(a)
Answer to Problem 21P
The magnetic field at center of loop is
Explanation of Solution
Given:
The radius of loop is
The current is
Formula used:
The expression for magnetic field is given by,
Calculation:
The magnetic field at origin is calculated as,
Further simplify the above,
Conclusion:
Therefore, the magnetic field at center of loop is
(b)
The magnitude of magnetic field at
(b)
Answer to Problem 21P
The magnetic field at center of loop is
Explanation of Solution
Calculation:
The magnetic field is calculated as,
Further simplify the above,
Conclusion:
Therefore, the magnetic field at center of loop is
(c)
The magnitude of magnetic field at
(c)
Answer to Problem 21P
The magnetic field at center of loop is
Explanation of Solution
Calculation:
The magnetic field is calculated as,
Further simplify the above,
Conclusion:
Therefore, the magnetic field at center of loop is
(d)
The magnitude of magnetic field at
(d)
Answer to Problem 21P
The magnetic field at center of loop is
Explanation of Solution
Calculation:
The magnetic field is calculated as,
Further simplify the above,
Conclusion:
Therefore, the magnetic field at center of loop is
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Chapter 27 Solutions
Physics for Scientists and Engineers
- 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 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_forwardAn ideal toroidal solenoid has an inner radius of 14.5 cm and outer radius of19.0 cm. It carries a current of 0.500 A and has 9500 number of turns. Find themagnetic field strengths at distances 15.0 cm and 20.0 cm from the center of the solenoid.arrow_forward
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- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning