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Problem 1: A 3.0-cm-diameter (Dc), 10-turn coil of wire, located at z = 0 in the xy-plane, carries a current I = 2.5 A. A 2.0-mm-diameter (D₁) conducting loop with R = 2.0 × 10-¹2 resistance is also in the xy-plane at the cen- ter of the coil. At t = 0s, the loop begins to move along the z-axis with a constant speed of v= 75 m/s. What is the induced current in the conducting loop at t = 200 µs? 10 turns ZA al FQI I FIG. 1: The scheme for Problem 1 a) The magnetic field created by the coil looks in general as shown in Fig. 1, but since the diameter of the loop is much smaller than that of the coil, we can use the formula for the on-axis magnetic field of IR² the coil. For a single turn of the coil, this formula gives B = 0. where Re is the radius of the coil. 2 (2²+R²)3/23 Derive the formula for the magnetic flux through the conducting loop as a function of the coordinate z. b) Derive the formula for the emf, E = |d, induced in the loop as it is moving up. To take the derivative of flux with respect to time, you need to use the chain rule for derivatives, d = do dz. What is the meaning of dễ in this formula, and how does z depend on time? Express & as a function of t. c) Using the formula for & from the previous step and the resistance of the loop, compute the value of the induced current at t = 200 µs. (Answer: loop = 4.4 x 10-² A)

Problem 1: A 3.0-cm-diameter (Dc), 10-turn coil of wire, located at z = 0 in the xy-plane, carries a current I = 2.5 A. A 2.0-mm-diameter (D₁) conducting loop with R = 2.0 × 10-¹2 resistance is also in the xy-plane at the cen- ter of the coil. At t = 0s, the loop begins to move along the z-axis with a constant speed of v= 75 m/s. What is the induced current in the conducting loop at t = 200 µs? 10 turns ZA al FQI I FIG. 1: The scheme for Problem 1 a) The magnetic field created by the coil looks in general as shown in Fig. 1, but since the diameter of the loop is much smaller than that of the coil, we can use the formula for the on-axis magnetic field of IR² the coil. For a single turn of the coil, this formula gives B = 0. where Re is the radius of the coil. 2 (2²+R²)3/23 Derive the formula for the magnetic flux through the conducting loop as a function of the coordinate z. b) Derive the formula for the emf, E = |d, induced in the loop as it is moving up. To take the derivative of flux with respect to time, you need to use the chain rule for derivatives, d = do dz. What is the meaning of dễ in this formula, and how does z depend on time? Express & as a function of t. c) Using the formula for & from the previous step and the resistance of the loop, compute the value of the induced current at t = 200 µs. (Answer: loop = 4.4 x 10-² 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, PART B, AND PART C BECAUSE I don't understand it is there any chance you can help me with the problems and can you label them as well

**Problem 1:** A 3.0-cm-diameter (\(D_c\)), 10-turn coil of wire, located at \(z = 0\) in the \(xy\)-plane, carries a current \(I = 2.5\) A. A 2.0-mm-diameter (\(D_l\)) conducting loop with \(R = 2.0 \times 10^{-4}\) \(\Omega\) resistance is also in the \(xy\)-plane at the center of the coil. At \(t = 0\) s, the loop begins to move along the \(z\)-axis with a constant speed of \(v = 75\) m/s. What is the induced current in the conducting loop at \(t = 200 \, \mu\)s? **Diagram Explanation:** *The diagram (Fig. 1) illustrates the setup for Problem 1. It shows a 10-turn coil lying flat on the \(xy\)-plane with the \(z\)-axis pointing upwards. The magnetic field \(B\) is depicted surrounding the coil, and the direction of current \(I\) is marked through the coil. The smaller conducting loop, initially at the center, moves along the +\(z\)-axis.* **a)** The magnetic field created by the coil looks in general as shown in Fig. 1, but since the diameter of the loop is much smaller than that of the coil, we can use the formula for the on-axis magnetic field of the coil. For a single turn of the coil, this formula gives \(B = \frac{\mu_0}{2} \frac{IR^2}{(z^2 + R_c^2)^{3/2}}\), where \(R_c\) is the radius of the coil. Derive the formula for the magnetic flux \(\Phi\) through the conducting loop as a function of the coordinate \(z\). **b)** Derive the formula for the emf, \(\mathcal{E} = \left| \frac{d\Phi}{dt} \right|\), induced in the loop as it is moving up. To take the derivative of flux with respect to time, you need to use the chain rule for derivatives, \(\frac{d\Phi}{dt} = \frac{d\Phi}{dz} \frac{dz}{dt}\
Transcribed Image Text:**Problem 1:** A 3.0-cm-diameter (\(D_c\)), 10-turn coil of wire, located at \(z = 0\) in the \(xy\)-plane, carries a current \(I = 2.5\) A. A 2.0-mm-diameter (\(D_l\)) conducting loop with \(R = 2.0 \times 10^{-4}\) \(\Omega\) resistance is also in the \(xy\)-plane at the center of the coil. At \(t = 0\) s, the loop begins to move along the \(z\)-axis with a constant speed of \(v = 75\) m/s. What is the induced current in the conducting loop at \(t = 200 \, \mu\)s? **Diagram Explanation:** *The diagram (Fig. 1) illustrates the setup for Problem 1. It shows a 10-turn coil lying flat on the \(xy\)-plane with the \(z\)-axis pointing upwards. The magnetic field \(B\) is depicted surrounding the coil, and the direction of current \(I\) is marked through the coil. The smaller conducting loop, initially at the center, moves along the +\(z\)-axis.* **a)** The magnetic field created by the coil looks in general as shown in Fig. 1, but since the diameter of the loop is much smaller than that of the coil, we can use the formula for the on-axis magnetic field of the coil. For a single turn of the coil, this formula gives \(B = \frac{\mu_0}{2} \frac{IR^2}{(z^2 + R_c^2)^{3/2}}\), where \(R_c\) is the radius of the coil. Derive the formula for the magnetic flux \(\Phi\) through the conducting loop as a function of the coordinate \(z\). **b)** Derive the formula for the emf, \(\mathcal{E} = \left| \frac{d\Phi}{dt} \right|\), induced in the loop as it is moving up. To take the derivative of flux with respect to time, you need to use the chain rule for derivatives, \(\frac{d\Phi}{dt} = \frac{d\Phi}{dz} \frac{dz}{dt}\
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