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Ch 30: Electromagnetic Induction
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 30: Electromagnetic InductionProblema 62
Capitolo 30, Problema 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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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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Faraday's Law

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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Magnetic Flux

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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Ballistic Pendulum Problems
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