![Physics for Scientists and Engineers, Technology Update (No access codes included)](https://www.bartleby.com/isbn_cover_images/9781305116399/9781305116399_largeCoverImage.gif)
In Figure P28.28, the cube is 40.0 cm on each edge. Four straight segments of wire—ab, bc, cd, and da—form a closed loop that carries a current I = 5.00 A in the direction shown. A uniform magnetic field of magnitude B = 0.020 0 T is in the positive y direction. Determine the magnetic force vector on (a) ab, (b) bc, (c) cd, and (d) da. (c) Explain how you could find the force exerted on the fourth of these segments from the forces on the other three, without further calculation involving the magnetic field.
Figure P28.28
(a)
![Check Mark](/static/check-mark.png)
Answer to Problem 29.44P
Explanation of Solution
Given info: The length of each edge of the cube is
The formula for the magnetic force is,
Here,
As the current flows down and the magnetic from point
Substitute
The magnitude of the force is zero because an equal and opposite force cancels it due to which the magnetic force of
Conclusion:
Therefore, magnetic force vector on
(b)
![Check Mark](/static/check-mark.png)
Answer to Problem 29.44P
Explanation of Solution
Given info: The length of each edge of the cube is
The formula for the magnetic force is,
Here,
As the current flows down and the magnetic from point
Substitute
By using the Flemings right hand rule, the thumb points towards the x axis direction. Thus, the direction of the magnetic force is in
Conclusion:
Therefore, the magnetic force vector on
(c)
![Check Mark](/static/check-mark.png)
Answer to Problem 29.44P
Explanation of Solution
Given info: The length of each edge of the cube is
The formula for the magnetic force is,
Here,
Use Pythagoras theorem to find
Substitute
As the current flows at an angle and the magnetic from point b is perpendicular so the angle
Substitute
By using the Flemings right hand rule, the thumb points towards the z axis direction. Thus, the direction of the magnetic force is in z direction.
Conclusion:
Therefore, the magnetic force vector on
(d)
![Check Mark](/static/check-mark.png)
Answer to Problem 29.44P
Explanation of Solution
Given info: The length of each edge of the cube is
The formula for the magnetic force is,
Here,
Use Pythagoras theorem to find
Substitute
As the current flows vertically and the magnetic from point b is perpendicular so the angle
Substitute
By using the Flemings right hand rule, the thumb points towards the direction d
The direction of the force is,
Thus, the direction of the magnetic force is
Conclusion:
Therefore, the magnetic force vector on
(e)
![Check Mark](/static/check-mark.png)
Answer to Problem 29.44P
Explanation of Solution
Given info: The length of each edge of the cube is
By the parallelogram law of forces, when the forces on three of the arms of a parallelogram are provided then the magnitude of force on the forth arm is equal to the resultant of the other three forces.
According to the parallelogram law of vectors,
Here,
Conclusion:
Therefore, the force exerted on the forth segment from the forces on the other three can be calculated by the parallelogram law of vectors.
Want to see more full solutions like this?
Chapter 29 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
- In Figure P22.43, the current in the long, straight wire is I1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries a current I2 = 10.0 A. The dimensions in the figure are c = 0.100 m, a = 0.150 m, and = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. Figure P22.43 Problems 43 and 44.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_forwardAcircularcoiofwireofradius5.Ocmhas2Otums and carries a current of 2.0 A. The coil lies in a magnetic field of magnitude 0.50 T that is directed parallel to the plane of the coil. (a) What is the magnetic dipole moment of the coil? (b) What is the torque on the coil?arrow_forward
- 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 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_forwardIn Niels Bohr’s 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29 × 10−11 m and its speed is 2.19 × 106 m/s. (a) What is the magnitude of the magnetic moment due to the electron’s motion? (b) If the electron moves in a horizontal circle, counterclockwise as seen from above, what is the direction of this magnetic moment vector?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_forwardFigure CQ19.7 shows a coaxial cable carrying current I in its inner conductor and a return current of the same magnitude in the opposite direction in the outer conductor. The magnetic field strength at r = r0 is Find the ratio B/B0, at (a) r = 2r0 and (b) r = 4r0. Figure CQ19.7arrow_forwardIn Figure P22.20, the cube is 40.0 cm on each edge. Four straight segments of wire—ab, bc, cd, and da—form a closed loop that carries a current I = 5.00 A in the direction shown. A uniform magnetic field of magnitude B = 0.020 0 T is in the positive y direction. Determine the magnetic force vector on (a) ab, (b) bc, (c) cd, and (d) da. (e) Explain how you could find the force exerted on the fourth of these segments from the forces on the other three, without further calculation involving the magnetic field.arrow_forward
- A magnetic field exerts a torque on each of the current carrying single loops of wire shown in Figure OQ22.12. The loops lie in the xy plane, each carrying the same magnitude current, and the uniform magnetic field points in the positive x direction. Rank the loops by the magnitude of the torque exerted on them by the field from largest to smallest Figure OQ22.12arrow_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_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
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781133104261/9781133104261_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781133939146/9781133939146_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305952300/9781305952300_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781285737027/9781285737027_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337553292/9781337553292_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781938168161/9781938168161_smallCoverImage.gif)