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Problem 2: A 2.0 cm x 2.0 cm square loop of wire with resistance R = 0.012 has one edge parallel to a long straight wire. The near edge of the loop is s = 1.0 cm from the wire (see the scheme in Fig. 2). I The current in the wire is increasing at the rate of d = 100 A/s. What is the current in the loop? a) The scheme in Fig. 2 illustrates the magnetic field created by the long straight wire at the location of the square loop. In the chosen coordinate system, the formula for the magnetic field can be written Hol 2πx' as B = Since the value of the magnetic field is not the same in all points across the square loop, the magnetic flux through the loop has to be found by taking the integral = ₂B da, where da is an infinitesimal patch of the area of the loop, da = dx dz, as shown in the scheme¹. Derive the formula for by taking this integral (you need to figure out the limits of integration on your own). T S Yo dz dx da L=2cm FIG. 2: The scheme for Problem 2 b) Derive the formula for the emf, & = |d. From & and the resistance of the loop compute the value of the induced current. (Answer: Iloop = 4.4 x 10-5 A).

Problem 2: A 2.0 cm x 2.0 cm square loop of wire with resistance R = 0.012 has one edge parallel to a long straight wire. The near edge of the loop is s = 1.0 cm from the wire (see the scheme in Fig. 2). I The current in the wire is increasing at the rate of d = 100 A/s. What is the current in the loop? a) The scheme in Fig. 2 illustrates the magnetic field created by the long straight wire at the location of the square loop. In the chosen coordinate system, the formula for the magnetic field can be written Hol 2πx' as B = Since the value of the magnetic field is not the same in all points across the square loop, the magnetic flux through the loop has to be found by taking the integral = ₂B da, where da is an infinitesimal patch of the area of the loop, da = dx dz, as shown in the scheme¹. Derive the formula for by taking this integral (you need to figure out the limits of integration on your own). T S Yo dz dx da L=2cm FIG. 2: The scheme for Problem 2 b) Derive the formula for the emf, & = |d. From & and the resistance of the loop compute the value of the induced current. (Answer: Iloop = 4.4 x 10-5 A).

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Hello, I really need help with part A and Part B because I don't understand it and I don't know what to do is there any chance you can help me with this problem and can you label it as well. thank you so much 

Problem 2: A 2.0 cm x 2.0 cm square loop of wire with resistance R = 0.0192 has one edge parallel to a long straight wire. The near edge of the loop is s = 1.0 cm from the wire (see the scheme in Fig. 2). I The current in the wire is increasing at the rate of = 100 A/s. What is the current in the loop? 2πx a) The scheme in Fig. 2 illustrates the magnetic field created by the long straight wire at the location of the square loop. In the chosen coordinate system, the formula for the magnetic field can be written as B = Ho Since the value of the magnetic field is not the same in all points across the square loop, the magnetic flux through the loop has to be found by taking the integral = ₂B da, where da is an infinitesimal patch of the area of the loop, da = dx dz, as shown in the scheme¹. Derive the formula for by taking this integral (you need to figure out the limits of integration on your own). vi YO dz dx da L=2cm x FIG. 2: The scheme for Problem 2 b) Derive the formula for the emf, & = |d. From & and the resistance of the loop compute the value of the induced current. (Answer: loop = 4.4 × 10-5 A).
Transcribed Image Text:Problem 2: A 2.0 cm x 2.0 cm square loop of wire with resistance R = 0.0192 has one edge parallel to a long straight wire. The near edge of the loop is s = 1.0 cm from the wire (see the scheme in Fig. 2). I The current in the wire is increasing at the rate of = 100 A/s. What is the current in the loop? 2πx a) The scheme in Fig. 2 illustrates the magnetic field created by the long straight wire at the location of the square loop. In the chosen coordinate system, the formula for the magnetic field can be written as B = Ho Since the value of the magnetic field is not the same in all points across the square loop, the magnetic flux through the loop has to be found by taking the integral = ₂B da, where da is an infinitesimal patch of the area of the loop, da = dx dz, as shown in the scheme¹. Derive the formula for by taking this integral (you need to figure out the limits of integration on your own). vi YO dz dx da L=2cm x FIG. 2: The scheme for Problem 2 b) Derive the formula for the emf, & = |d. From & and the resistance of the loop compute the value of the induced current. (Answer: loop = 4.4 × 10-5 A).
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