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Ch 30: Electromagnetic Induction
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 30: Electromagnetic InductionProblem 62
Chapter 30, Problem 62

One way to measure the strength of a magnetic field is with a flip coil. Suppose a 200-turn, 4.0-cm-diameter coil with a resistance of 2.0 Ω is connected to a ballistic galvanometer, a device that measures the total charge passing through. The coil is held perpendicular to the field, then quickly flipped 180° so that the opposite side is facing the magnetic field. Afterward, the galvanometer reads 7.5 μC. What is the field strength? Hint: Use I = dq/dt to relate the net change of flux to the amount of charge that flows through the galvanometer.

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Step 1: Start by understanding the relationship between the change in magnetic flux and the charge passing through the galvanometer. The total charge (q) is related to the electromotive force (EMF) induced in the coil and the resistance (R) of the coil using Ohm's Law: \( \text{EMF} = IR \), where \( I = \frac{dq}{dt} \).
Step 2: The EMF induced in the coil is also related to the rate of change of magnetic flux through Faraday's Law of Induction: \( \text{EMF} = -N \frac{d\Phi_B}{dt} \), where \( N \) is the number of turns in the coil and \( \Phi_B \) is the magnetic flux. Since the coil is flipped 180°, the change in flux is \( \Delta \Phi_B = 2 \Phi_B \).
Step 3: The magnetic flux \( \Phi_B \) through the coil is given by \( \Phi_B = B A \cos \theta \), where \( B \) is the magnetic field strength, \( A \) is the area of the coil, and \( \theta \) is the angle between the field and the normal to the coil. Initially, \( \cos \theta = 1 \) (perpendicular to the field), and after flipping, \( \cos \theta = -1 \). Thus, \( \Delta \Phi_B = 2 B A \).
Step 4: Substitute \( \Delta \Phi_B \) into Faraday's Law: \( \text{EMF} = -N \frac{\Delta \Phi_B}{\Delta t} \). Since \( \Delta t \) is not explicitly given, we use the relationship \( \text{EMF} = IR \) and \( I = \frac{q}{\Delta t} \) to eliminate \( \Delta t \). This gives \( \text{EMF} = \frac{q R}{N \Delta \Phi_B} \).
Step 5: Solve for the magnetic field strength \( B \). Rearrange the equation \( \Delta \Phi_B = 2 B A \) to \( B = \frac{\Delta \Phi_B}{2 A} \). Substitute \( A = \pi r^2 \) (where \( r \) is the radius of the coil) and use the given values: \( N = 200 \), \( R = 2.0 \ \Omega \), \( q = 7.5 \ \mu C \), and the diameter of the coil to find \( r \).

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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 and the number of turns in the coil. This principle is fundamental in understanding how the flipping of the coil in a magnetic field generates a measurable current, which is crucial for solving the problem.
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Magnetic Flux

Magnetic flux is defined as the product of the magnetic field strength and the area through which the field lines pass, taking into account the angle between the field and the normal to the surface. It is measured in Weber (Wb) and is essential for calculating the change in flux when the coil is flipped. The change in magnetic flux is what induces the current measured by the galvanometer.
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Ballistic Galvanometer

A ballistic galvanometer is a sensitive instrument used to measure the total charge that passes through it in a short time interval. It operates on the principle of electromagnetic induction, where the deflection of the galvanometer needle is proportional to the charge that flows. In this problem, the reading of 7.5 μC provides the necessary data to relate the induced EMF to the magnetic field strength.
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Related Practice
Textbook Question

BIO One possible concern with MRI (see Exercise 28) is turning the magnetic field on or off too quickly. Bodily fluids are conductors, and a changing magnetic field could cause electric currents to flow through the patient. Suppose a typical patient has a maximum cross-section area of 0.060 m2. What is the smallest time interval in which a 5.0 T magnetic field can be turned on or off if the induced emf around the patient's body must be kept to less than 0.10 V?

Textbook Question

A 50 cm solenoid with 1000 turns has an inductance of 20 mH. What is the magnetic field strength inside the inductor when the current is 75 mA?

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Textbook Question

INT You've decided to make the magnetic projectile launcher shown in FIGURE P30.58 for your science project. An aluminum bar slides along metal rails through a magnetic field B. The switch closes at t = 0 s, while the bar is at rest, and a battery of emf εbat starts a current flowing around the loop. The battery has internal resistance r. The resistances of the rails, which are separated by distance l, and the bar are effectively zero. Evaluate vterm for εbat = 1.0 V, r = 0.10 Ω, l = 6.0 cm, and B = 0.50 T.

Textbook Question

INT FIGURE P30.59 shows a U-shaped conducting rail that is oriented vertically in a horizontal magnetic field. The rail has no electric resistance and does not move. A slide wire with mass m and resistance R can slide up and down without friction while maintaining electrical contact with the rail. The slide wire is released from rest. Determine the value of vterm if l = 20 cm,m = 10 g, R = 0.10 Ω, and B = 0.50 T.

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Textbook Question

CALC FIGURE P30.67 shows the potential difference across a 20 mH inductor. The current through the inductor at t = 0 ms is 0.25 A. What is the current at t = 10 ms?

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Textbook Question

INT You've decided to make the magnetic projectile launcher shown in FIGURE P30.58 for your science project. An aluminum bar slides along metal rails through a magnetic field B. The switch closes at t = 0 s, while the bar is at rest, and a battery of emf εbat starts a current flowing around the loop. The battery has internal resistance r. The resistances of the rails, which are separated by distance l, and the bar are effectively zero. Show that the bar reaches a terminal speed vterm, and find an expression for vterm.