/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A flat coil of wire has an area ... [FREE SOLUTION] | 91Ó°ÊÓ

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A flat coil of wire has an area \(A, N\) turns, and a resistance \(R .\) It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of \(90^{\circ},\) so that the normal becomes perpendicular to the magnetic field. The coil has an area of \(1.5 \times 10^{-3} \mathrm{m}^{2}, 50\) turns, and a resistance of 140\(\Omega .\) During the time while it is rotating, a charge of \(8.5 \times 10^{-5}\) C flows in the coil. What is the magnitude of the magnetic field?

Short Answer

Expert verified
The magnetic field magnitude is 0.1587 T.

Step by step solution

01

Calculate EMF generated in the coil

The charge that flows through the coil is given as \(8.5 \times 10^{-5}\) C. To find the EMF induced in the coil, we can use the formula \( Q = \frac{\text{EMF}}{R} \), where \( Q \) is the charge, and \( R \) is the resistance. Rearranging gives us \( \text{EMF} = Q \times R = 8.5 \times 10^{-5} \times 140 = 0.0119 \, \text{V} \).
02

Relate EMF to change in magnetic flux

The EMF induced in the coil can also be expressed in terms of the change in magnetic flux: \( \text{EMF} = N \frac{\Delta \Phi}{\Delta t} \), where \( N \) is the number of turns and \( \Delta \Phi \) is the change in magnetic flux. Since the coil rotates from parallel to perpendicular to the field, the change in magnetic flux is \( N \cdot A \cdot B \).
03

Calculate the magnetic field

We have the expression for EMF as \( 0.0119 = N A B \). Plugging in the values, \( 0.0119 = 50 \times (1.5 \times 10^{-3}) \times B \). Solve for \( B \): \[ B = \frac{0.0119}{50 \times 1.5 \times 10^{-3}} = \frac{0.0119}{0.075} = 0.1587 \, \text{T} \]
04

Conclusion

The magnitude of the magnetic field is calculated by the formula derived, taking into account all physical properties given in the problem. The correct expression and substitution result in the final value for the magnetic field's magnitude.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Flux
Magnetic flux is a fundamental concept in electromagnetism that describes the total magnetic field passing through a specified area. This concept is essential to understanding electromagnetic induction. Magnetic flux, denoted as \( \Phi \), is calculated by the equation \( \Phi = B \cdot A \cdot \cos(\theta) \). Here, \( B \) is the magnetic flux density, \( A \) is the area through which the field lines pass, and \( \theta \) is the angle between the field lines and the normal (perpendicular) to the surface.

In the context of the given exercise, the coil initially has its normal parallel to the magnetic field. As the coil rotates to a position perpendicular to the magnetic field, the angle \( \theta \) changes from 0° to 90°. This change alters the magnetic flux through the coil because the component of the magnetic field passing through the area diminishes.

The change in magnetic flux is directly related to the induced electromotive force (EMF), as captured by Faraday's Law of Induction.
EMF (Electromotive Force)
Electromotive Force, commonly abbreviated as EMF, is a measure of the energy provided by a source of electrical current. Although it's called a 'force,' it is more accurately described as a potential difference or voltage. When we speak of EMF in electromagnetic induction, we refer to the voltage generated by a changing magnetic field.

In the scenario of the coil rotating in a magnetic field, the EMF is generated because the coil cuts through magnetic field lines. According to the formula derived in the step-by-step solution, the EMF induced can be determined by the relationship \( Q = \frac{\text{EMF}}{R} \), where \( Q \) is the electrical charge passed through the circuit, and \( R \) is the coil's resistance.

This relationship highlights how the movement within the field results in a measurable potential difference across the coil, allowing us to calculate the strength of EMF (expressed in volts) induced by the changes in magnetic flux.
Faraday's Law of Induction
Faraday's Law of Induction is a key principle in electromagnetism that quantifies how a change in magnetic flux induces an electromotive force (EMF) in a closed circuit. This law is mathematically expressed as \( \text{EMF} = -N \frac{\Delta \Phi}{\Delta t} \), where \( N \) is the number of turns in the coil, \( \Delta \Phi \) is the change in magnetic flux, and \( \Delta t \) is the time interval over which the change occurs.

This law underpins the behavior observed in the exercise, where the coil's rotation through 90 degrees changes the magnetic flux through the coil. Consequently, an EMF is induced, causing a current to flow. The negative sign in the equation represents Lenz's Law, which states that the induced EMF will always oppose the change in flux, striving to counteract it.

By applying Faraday's Law, one can understand how variations in flux lead to the production of electrical currents, crucial for technologies like electric generators and transformers.
Resistance in Electrical Circuits
Resistance is the measure of how strongly an object opposes the flow of electric charge. Its unit is the ohm (\( \Omega \)). In the context of the rotating coil, resistance plays a critical role in determining how much current flows when EMF is induced.

The relationship between resistance, EMF, and current is given by Ohm's Law: \( V = IR \), where \( V \) is voltage (or EMF in this case), \( I \) is current, and \( R \) is resistance. This law is visible in the step-by-step solution, where the EMF is calculated from the flowing charge and the coil's resistance: \( \text{EMF} = Q \times R \).

Understanding resistance is vital to predicting how much induced current will appear for a given EMF, and it shows how energy loss to heat is inevitable when current flows through the resistance of a conductor. In practical applications, managing resistance ensures efficient energy use.

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Most popular questions from this chapter

The resistances of the primary and secondary coils of a transformer are 56 and \(14 \Omega,\) respectively. Both coils are made from length of the same copper wire. The circular turns of each coil have the same diameter. Find the turns ratio \(N_{s} / N_{p} .\)

ssm The plane of a flat, circular loop of wire is horizontal. An external magnetic field is directed perpendicular to the plane of the loop. The magnitude of the external magnetic field is increasing with time. Because of this increasing magnetic field, an induced current is flowing clockwise in the loop, as viewed from above. What is the direction of the external magnetic field? Justify your conclusion.

ssm Multiple-Concept Example 13 reviews some of the principles used in this problem. Suppose you wish to make a solenoid whose self-inductance is 1.4 \(\mathrm{mH}\) . The inductor is to have a cross-sectional area of \(1.2 \times 10^{-3} \mathrm{m}^{2}\) and a length of 0.052 \(\mathrm{m}\) . How many turns of wire are needed?

The coil of a generator has a radius of 0.14 \(\mathrm{m}\) . When this coil is unwound, the wire from which it is made has a length of 5.7 \(\mathrm{m}\) . The magnetic field of the generator is \(0.20 \mathrm{T},\) and the coil rotates at an angular speed of 25 \(\mathrm{rad} / \mathrm{s}\) . What is the peak emf of this generator?

Parts \(a\) and \(b\) of the drawing show the same uniform and constant (in time) magnetic field \(\overrightarrow{\mathbf{B}}\) directed perpendicularly into the paper over a rectangular region. Outside this region, there is no field. Also shown is a rectangular coil (one turn), which lies in the plane of the paper. In part \(a\) the long side of the coil (length \(=L )\) is just at the edge of the field region, while in part \(b\) the short side (width \(=W\) is just at the edge. It is known that \(L / W=3.0 .\) In both parts of the drawing the coil is pushed into the field with the same velocity \(\overrightarrow{\mathbf{v}}\) until it is completely within the field region. The magnitude of the average emf induced in the coil in part \(a\) is 0.15 V. What is its magnitude in part \(b\) ?

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