Concept explainers
(a)
The magnitude and the direction of the net magnetic field at the mid way between the wires.
(a)
Answer to Problem 19P
The magnitude of the net magnetic field is
Explanation of Solution
Write the expression to obtain the magnetic field along the conductor.
Here,
Write the expression to obtain the magnetic field due to the wire
Here,
Write the expression to obtain the magnetic field due to the wire
Here,
Write the expression to obtain the net magnetic field at the mid way between the wires.
Here,
Substitute
Further substitute
Conclusion:
Substitute
Therefore, the magnitude of the net magnetic field is
(b)
The magnitude and the direction of the net magnetic field at point
(b)
Answer to Problem 19P
The magnitude of the net magnetic field is
Explanation of Solution
Write the expression to obtain the magnetic field due to the wire
Here,
Write the expression to obtain the magnetic field due to the wire
Here,
Write the expression to obtain the net magnetic field at the mid way between the wires.
Here,
Substitute
Further substitute
Conclusion:
Substitute
Therefore, the magnitude of the net magnetic field is
(c)
The magnitude and the direction of the net magnetic field at point
(c)
Answer to Problem 19P
The magnitude of the net magnetic field is
Explanation of Solution
Write the expression to obtain the magnetic field due to the wire
Here,
Write the expression to obtain the magnetic field due to the wire
Here,
Write the expression to obtain the net magnetic field at the mid way between the wires.
Here,
Substitute
Further substitute
Conclusion:
Substitute
Therefore, the magnitude of the net magnetic field is
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Chapter 30 Solutions
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
- Two infinitely long current-carrying wires run parallel in the xy plane and are each a distance d = 11.0 cm from the y axis (Fig. P30.83). The current in both wires is I = 5.00 A in the negative y direction. a. Draw a sketch of the magnetic field pattern in the xz plane due to the two wires. What is the magnitude of the magnetic field due to the two wires b. at the origin and c. as a function of z along the z axis, at x = y = 0? FIGURE P30.83arrow_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 strong magnet is placed under a horizontal conducting ring of radius r that carries current I as shown in Figure P28.27. If the magnetic field B makes an angle with the vertical at the rings location, what are (a) the magnitude and (b) the direction of the resultant magnetic force on the ring? Figure P28.27arrow_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 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_forwardSketch a plot of the magnitude of the magnetic field as a function of position r for a coax (Fig. P31.27).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_forwardFigure P30.10 shows a circular current-carrying wire. Using the coordinate system indicated (with the z axis out of the page), state the direction of the magnetic field at points A and B.arrow_forward
- Determine 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_forwardWhy 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_forward
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