Concept explainers
(a)
The magnitude and direction of magnetic field at point
(a)
Answer to Problem 7P
The magnitude of magnetic field at point
Explanation of Solution
Given Info: The current flowing through the conductor is
Diagram of three parallel conductor having current of magnitude
Figure (1)
Formula to calculate side of the square is,
Formula to calculate angle
Formula to calculate magnetic field at point
Here
Substitute
Formula to calculate magnetic field at point
Here
Substitute
Formula to calculate magnetic field at point
Here
Substitute
Write the expression to calculate magnetic field at point
Here,
Substitute
substitute
Write the expression to calculate y-component of magnetic field at
Substitute
Formula to calculate net magnetic field at point
Substitute
Hence, magnetic field at point
Conclusion:
Therefore magnetic field at point
(b)
magnitude and direction of magnetic field at point
(b)
Answer to Problem 7P
magnitude of magnetic field at point
Explanation of Solution
Formula to calculate magnetic field at point
Here
Substitute
Formula to calculate magnetic field at point
Here
Substitute
Formula to calculate magnetic field at point
Here
Substitute
Write the expression to calculate magnetic field at point
Here,
Substitute
substitute
Write the expression to calculate y-component of magnetic field at
Substitute
Formula to calculate net magnetic field at point
Substitute
Hence, magnetic field at point
Conclusion:
Therefore magnetic field at point
(c)
magnitude and direction of magnetic field at point
(c)
Answer to Problem 7P
magnitude of magnetic field at point
Explanation of Solution
Formula to calculate magnetic field at point
Here
Substitute
Formula to calculate magnetic field at point
Here
Substitute
Formula to calculate magnetic field at point
Here
Substitute
Write the expression to calculate magnetic field at point
Here,
Substitute
Write the expression to calculate y-component of magnetic field at
Substitute
Formula to calculate net magnetic field at point
Substitute
Hence, magnetic field at point
Conclusion:
Therefore magnetic field at point
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Chapter 29 Solutions
Physics for Scientists and Engineers with Modern Physics
- 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_forwardIn 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_forwardThe velocity vector of a singly charged helium ion (mHe = 6.64 1027 kg) is given by v=4.50105m/s. The acceleration of the ion in a region of space with a uniform magnetic field is 8.50 1012 m/s2 in the positive y direction. The velocity is perpendicular to the field direction. What are the magnitude and direction of the magnetic field in this region?arrow_forward
- A uniform magnetic field of magnitude is directed parallel to the z-axis. A proton enters the field with a velocity v=(4j+3k)106m/s and travels in a helical path with a radius of 5.0 cm. (a) What is the value of B? (b) What is the time required for one trip around the helix? (c) Where is the proton 5.0107s after entering the field?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_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_forward
- A 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_forwardA wire carrying a current I is bent into the shape of an exponential spiral, r = e, from = 0 to = 2 as suggested in Figure P29.47. To complete a loop, the ends of the spiral are connected by a straight wire along the x axis. (a) The angle between a radial line and its tangent line at any point on a curve r = f() is related to the function by tan=rdr/d Use this fact to show that = /4. (b) Find the magnetic field at the origin. Figure P29.47arrow_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_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_forwardTwo long, straight wires are parallel and 25 cm apart. (a) If each wire carries a current of 50 A in the same direction, what is the magnetic force per meter exerted on each wire? (b) Does tire force pull the wires together or push them apart? (c) What happens if the currents flow in opposite directions?arrow_forwardAn electron in a TV CRT moves with a speed of 6.0107 m/s, in a direction perpendicular to Earth's field, which has a strength of 5.0105 T. (a) What strength electric field must be applied perpendicular to the Earth’s field to make the election moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a collection,)arrow_forward
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