Concept explainers
(a)
The current that is needed to generate a magnetic field that has a magnitude of 0.50 T
Answer to Problem 71QAP
The current that is needed to generate a magnetic field of 0.50 T= 3.0 x 102 A(rounded to two significant figures)
Explanation of Solution
Given:
A coil of wire has a diameter of 15 cm = 0.15 m is consisting 250 windings. The magnitude of the magnetic field generated by the coil due to a current that flows through is 0.050 T at a perpendicular distance of 30 cm from the center of the coil.
Formula used:
The z component of the magnetic field(Bz) for a coil of wire with N windings carrying a current i is given by the following equation.
Calculation:
It is given that the magnitude of the magnetic field at perpendicular distance 3.0 cm from the center of the coil is 0.50 T. The diameter of the coil is given as 15 cm which implies that the radius of the coil is 7.5 cm. Also, the number of the windings of the coil is given as 250.Substituing these values to equation (a) one could easily find the current that generate the desired magnetic field of 0.50 T.
Conclusion:
The current that is needed to generate a magnetic field of 0.50 T= 3.0 x 102 A (rounded to two significant figures)
(b)
The magnitude of magnetic field at the center of the coil at the forehead
Answer to Problem 71QAP
The magnitude of the magnetic field at the center of the forehead= 0.62 T
Explanation of Solution
Given:
A coil of wire has a diameter of 15 cm = 0.15 m is consisting 250 windings. The magnitude of the magnetic field generated by the coil due to a current that flows through is 0.050 T at a perpendicular distance of 30 cm from the center of the coil.
Formula used:
The z component of the magnetic field(Bz) for a coil of wire with N windings carrying a current i is given by the following equation.
Calculation:
It is deduced from part a) that the current that generate a magnetic field of magnitude 0.50 T at a perpendicular distance of 3.0 cm from the center of the coil is 298 A. Now we are asked to calculate the magnitude of the magnetic field at the center of the coil. The diameter of the coil is given as 15 cm which implies that the radius of the coil is 7.5 cm. Also, the number of the windings of the coil is given as 250.Substituing these values to equation (a) one could easily find the magnitude of the magnetic field at the center. Note here that since we have to calculate the magnetic field at the center of the coil the perpendicular distance measured from the center of the coil(z) is 0 cm.
Substituting to equation(a);
Conclusion:
The magnitude of the magnetic field at the center of the coil = 0.62 T
(c)
If the current needed in part a) is too high how could one easily achieve the same magnetic field of 0.50 T
Answer to Problem 71QAP
The easiest way that one could compensate for higher currents inside a loop in order to generate a particular magnetic field magnitude is to increase the number of windings in the loop.
Explanation of Solution
Given:
From the calculations in part a) it has been deduced that a current of 298 A is required to generate a magnetic field of 0.50 T from a loop that has 250 windings.
Calculation:
The z component of the magnetic field(Bz) for a coil of wire with N windings carrying a current i is given by the following equation.
Careful inspection of equation (a) reveals us that Bz is directly proportional to N.So instead of increasing the current(i) one could increase N to achieve higher Bz values.
Conclusion:
The easiest way that one could compensate for higher currents inside a loop in order to generate a particular magnetic field magnitude is to increase the number of windings in the loop
Want to see more full solutions like this?
Chapter 19 Solutions
COLLEGE PHYSICS
- A 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 toroid with an inner radius of 20 cm and an outer radius of 22 cm is tightly wound with one layer of wire that has a diameter of 0.25 mm. (a) How many turns are there on the toroid? (b) If the current through the toroid windings is 2.0 A, what is the strength of the magnetic field at the center of the toroid?arrow_forwardReview. In studies of the possibility of migrating birds using the Earths magnetic field for navigation, birds have been fitted with coils as caps and collars as shown in Figure P22.39. (a) If the identical coils have radii of 1.20 cm and are 2.20 cm apart, with 50 turns of wire apiece, what current should they both carry to produce a magnetic field of 4.50 105 T halfway between them? (b) If the resistance of each coil is 210 V, what voltage should the battery supplying each coil have? (c) What power is delivered to each coil? Figure P22.39arrow_forward
- A thin copper rod 1.00 m long has a mass of 50.0 g. What is the minimum current in the rod that would allow it to levitate above the ground in a magnetic field of magnitude 0.100 T? (a) 1.20 A (b) 2.40 A (c) 4.90 A (d) 9.80 A (e) none of those answersarrow_forwardA piece of insulated wire is shaped into a figure eight as shown in Figure P23.12. For simplicity, model the two halves of the figure eight as circles. The radius of the upper circle is 5.00 cm and that of the lower circle is 9.00 cm. The wire has a uniform resistance per unit length of 3.00 Ω/m. A uniform magnetic field is applied perpendicular to the plane of the two circles, in the direction shown. The magnetic field is increasing at a constant rate of 2.00 T/s. Find (a) the magnitude and (b) the direction of the induced current in the wire. Figure P23.12arrow_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
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning