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A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and lies in the xy plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop. (a) What is the direction of the magnetic force exerted on wire segment ab? (b) What is the direction of the torque associated with this force about an axis through the origin? (c) What is the direction of the magnetic force exerted on segment cd? (d) What is the direction of the torque associated with this force about an axis through the origin? (e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x axis? (f) Can they affect the motion of the loop in any way? Explain. (g) What is the direction of the magnetic force exerted on segment bc? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? (j) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x axis? (k) Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment sector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (l).
Figure P28.34
(a)
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Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.
By using the Fleming left hand rule, if the middle finger shows the direction of current in segment
Conclusion:
Therefore, the direction of the force is in the positive x-axis.
(b)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
The Formula to calculate the torque associated with the force is,
Here,
Substitute
The torque associated in the above expression in the negative of the unit vector
Conclusion:
Therefore, the direction of the torque associated with the force on segment
(c)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
The Formula to calculate the magnetic force on the wire segment
Here,
Substitute
In the above expression it is shown that the magnetic force is in the negative of the unit vector
Conclusion:
Therefore, the direction of the magnetic force exerted on the segment
(d)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
The Formula to calculate the torque associated with the force is,
Here,
Substitute
The torque associated in the above expression is in the negative of the unit vector
Conclusion:
Therefore, the direction of the torque associated with the force on segment
(e)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
As the force examined on the segment
Thus the net force on the rectangular loop become zero.
Conclusion:
Therefore, the magnetic force cannot rotate the loop.
(f)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem the magnetic field, the current and the length of the loop is constant, therefore, the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.
Conclusion:
Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.
(g)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
The Formula to calculate the magnetic force on the wire segment
Here,
Substitute
In the above expression it is shown that the magnetic force is in the negative of the unit vector
Conclusion:
Therefore, the magnetic force on segment
(h)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
The Formula to calculate the torque associated with the force is,
Here,
Substitute
The torque associated in the above expression is in the positive of the unit vector
Conclusion:
Therefore, the direction of the torque associated with the force on segment
(i)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
The Formula to calculate the torque associated with the force is,
The segment
Conclusion:
Therefore, the torque on the segment
(j)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
As the torque on the segment
Conclusion:
Therefore, the rectangular loop will rotate in the anticlockwise direction.
(k)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
The formula to calculate the magnetic moment of the loop is,
Here,
Substitute
Conclusion:
Therefore, the magnitude of the magnetic moment of the loop is
(l)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,
Conclusion:
Therefore, the angle between the magnetic moment and magnetic field is
(m)
![Check Mark](/static/check-mark.png)
Answer to Problem 34P
Explanation of Solution
Given Info: The length of the rectangular loop is
Formula to calculate the torque in current carrying wire is,
Substitute
Conclusion:
Therefore, the torque on the loop is
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Chapter 28 Solutions
Physics for Scientists and Engineers
- 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 circular coil 15.0 cm in radius and composed of 145 tightly wound turns carries a current of 2.50 A in the counterclockwise direction, where the plane of the coil makes an angle of 15.0 with the y axis (Fig. P30.73). The coil is free to rotate about the z axis and is placed in a region with a uniform magnetic field given by B=1.35jT. a. What is the magnitude of the magnetic torque on the coil? b. In what direction will the coil rotate? FIGURE P30.73arrow_forwardA wire is bent in the form of a square loop with sides of length L (Fig. P30.24). If a steady current I flows in the loop, determine the magnitude of the magnetic field at point P in the center of the square. FIGURE P30.24arrow_forward
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- Why is the following situation impossible? Figure P28.46 shows an experimental technique for altering the direction of travel for a charged particle. A particle of charge q = 1.00 C and mass m = 2.00 1015 kg enters the bottom of the region of uniform magnetic field at speed = 2.00 105 m/s, with a velocity vector perpendicular to the field lines. The magnetic force on the particle causes its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an angle from its original direction. The magnetic field has magnitude B = 0.400 T and is directed out of the page. The length h of the magnetic field region is 0.110 m. An experimenter performs the technique and measures the angle at which the particles exit the top of the field. She finds that the angles of deviation are exactly as predicted. Figure P28.46arrow_forwardA toroid has a major radius R and a minor radius r and is tightly wound with N turns of wire on a hollow cardboard torus. Figure P31.6 shows half of this toroid, allowing us to see its cross section. If R r, the magnetic field in the region enclosed by the wire is essentially the same as the magnetic field of a solenoid that has been bent into a large circle of radius R. Modeling the field as the uniform field of a long solenoid, show that the inductance of such a toroid is approximately L=120N2r2R Figure P31.6arrow_forwardA metal rod of mass m slides without friction along two parallel horizontal rails, separated by a distance and connected by a resistor R, as shown in Figure P30.13. A uniform vertical magnetic field of magnitude B is applied perpendicular to the plane of the paper. The applied force shown in the figure acts only for a moment, to give the rod a speed v. In terms of m, , R, B, and v, find the distance the rod will then slide as it coasts to a stop. Figure P30.13arrow_forward
- Figure 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_forwardA rectangular coil consists of N = 100 closely wrapped turns and has dimensions a = 0.400 m and b = 0.300 m. The coil is hinged along the y axis, and its plane makes an angle = 30.0 with the x axis (Fig. P22.25). (a) What is the magnitude of the torque exerted on the coil by a uniform magnetic field B = 0.800 T directed in the positive x direction when the current is I = 1.20 A in the direction shown? (b) What is the expected direction of rotation of the coil? Figure P22.25arrow_forwardDetermine the initial direction of the deflection of charged particles as they enter the magnetic fields as shown in Figure P22.2. Figure P22.2.arrow_forward
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