Concept explainers
(a)
The magnitude of the magnetic field created by
(a)
Answer to Problem 45P
The magnitude of the magnetic field created by
Explanation of Solution
Write the expression of the magnetic field at the origin due to the current
Here,
Conclusion:
Substitute
The magnitude of the magnetic field created by
(b)
The force per unit length
(b)
Answer to Problem 45P
The force per unit length exerted by
Explanation of Solution
Write the expression of the force exerted per unit length by
Here,
Conclusion:
Substitute
The force per unit length exerted by
(c)
The magnitude of the magnetic field created by
(c)
Answer to Problem 45P
The magnitude of the magnetic field created by
Explanation of Solution
Write the expression of the magnetic field at the origin due to the current
Here,
Conclusion:
Substitute
The magnitude of the magnetic field created by
(d)
The force per unit length
(d)
Answer to Problem 45P
The force per unit length exerted by
Explanation of Solution
Write the expression of the force exerted per unit length by
Here,
Conclusion:
Substitute
The force per unit length exerted by
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Chapter 22 Solutions
Principles of Physics
- 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 wire 2.80 m in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) 60.0, (b) 90.0, and (c) 120.arrow_forwardTwo long, straight, parallel wires carry currents that are directed perpendicular to the page as shown in Figure P30.9. Wire 1 carries a current I1, into the page (in the negative z direction) and passes through the x axis at x = +. Wire 2 passes through the x axis at x = 2a and carries an unknown current I2. The total magnetic field at the origin due to the current-carrying wires has the magnitude 20I1(2a). The current I2 can have either of two possible values, (a) Find the value of with the smaller magnitude, stating it in terms of I1, and giving its direction. (b) Find the other possible value of I2.arrow_forward
- Within the green dashed circle shown in Figure P23.28, 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 P1, which is at a distance r1 = 5.00 cm from the center of the circular field region. (c) At what instant is this force equal to zero?arrow_forwardTwo long, parallel wires carry currents of I1 = 3.00 A and I2 = 5.00 A in the directions indicated in Figure P29.11 (page 792). (a) Find the magnitude and direction of the magnetic field at a point midway between the wires. (b) Find the magnitude and direction of the magnetic field at point P, located d = 20.0 cm above the wire carrying the 5.00-A current. Figure P29.11arrow_forwardOne 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_forward
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