Textbook Question
FIGURE EX30.13 shows a 10-cm-diameter loop in three different magnetic fields. The loop's resistance is 0.20 Ω. For each, what are the size and direction of the induced current?
Ch 30: Electromagnetic Induction
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 30, Problem 17b
CALC A 5.0-cm-diameter coil has 20 turns and a resistance of 0.50 Ω. A magnetic field perpendicular to the coil is B = 0.020t + 0.010t2, where B is in tesla and t is in seconds. Evaluate I at t = 5 s and t = 10 s.
Verified step by step guidance1
Step 1: Calculate the area of the coil. The coil has a diameter of 5.0 cm, so the radius is 2.5 cm or 0.025 m. The area of the coil is given by the formula A = πr². Substitute the radius into the formula to find the area.
Step 2: Determine the magnetic flux through the coil. Magnetic flux (Φ) is given by Φ = N * B * A, where N is the number of turns, B is the magnetic field, and A is the area of the coil. Substitute the values for N, B, and A into the formula. Note that B is a function of time, B = 0.020t + 0.010t².
Step 3: Find the rate of change of magnetic flux (dΦ/dt). Differentiate the expression for Φ with respect to time t. Use the product rule if necessary, as B is time-dependent.
Step 4: Calculate the induced electromotive force (EMF) using Faraday's law of induction. Faraday's law states that EMF = -dΦ/dt. Substitute the result from Step 3 into this formula.
Step 5: Determine the current (I) in the coil using Ohm's law. Ohm's law states that I = EMF / R, where R is the resistance of the coil. Substitute the values for EMF and R to find the current at t = 5 s and t = 10 s.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
7mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Faraday's Law of Electromagnetic Induction
Faraday's Law states that a change in magnetic flux through a coil induces an electromotive force (EMF) in the coil. The induced EMF is proportional to the rate of change of the magnetic flux, which can be calculated using the formula EMF = -dΦ/dt, where Φ is the magnetic flux. This principle is fundamental for understanding how coils generate current when exposed to changing magnetic fields.
Recommended video:
Guided course
09:26
Faraday's Law
Ohm's Law
Ohm's Law relates the voltage (V), current (I), and resistance (R) in an electrical circuit, expressed as V = IR. This law is crucial for calculating the current flowing through the coil once the induced EMF is known. By rearranging the formula, we can find the current as I = V/R, allowing us to determine how the resistance of the coil affects the current generated by the induced EMF.
Recommended video:
Guided course
03:07Resistance and Ohm's Law
Magnetic Flux
Magnetic flux (Φ) is defined as the product of the magnetic field (B) and the area (A) through which it passes, taking into account the angle (θ) between the field lines and the normal to the surface: Φ = B·A·cos(θ). In this problem, the magnetic field varies with time, which affects the flux through the coil and is essential for calculating the induced EMF and resulting current at specific times.
Recommended video:
Guided course
04:52Magnetic Flux
Related Practice
Textbook Question
CALC A 5.0-cm-diameter coil has 20 turns and a resistance of 0.50 Ω. A magnetic field perpendicular to the coil is B = 0.020t + 0.010t2, where B is in tesla and t is in seconds. Find an expression for the induced current I(t) as a function of time.
Textbook Question
FIGURE EX30.19 shows the current as a function of time through a 20-cm-long, 4.0-cm-diameter solenoid with 400 turns. Draw a graph of the induced electric field strength as a function of time at a point 1.0 cm from the axis of the solenoid.
Textbook Question
A 12-cm-diameter, 1.0-m-long solenoid is wound with 2000 turns of superconducting wire. When the magnet is turned on, the current increases from 0 to Imax in 2.5 s. At t = 1.0 s, the induced electric field midway between the axis and the windings is 7.5×10−3 V/m. What is the solenoid's steady magnetic field strength?
Textbook Question
A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of the coil is parallel to the field. What is the emf of the coil?
Textbook Question
The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.0 T/s. What is the electric field strength inside the solenoid at a point (a) on the axis and (b) 2.0 cm from the axis?
1
views