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# Chapter 29--Examples PowerPoint PPT Presentation

Chapter 29--Examples. Problem.

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Chapter 29--Examples

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## Chapter 29--Examples

### Problem

• Derive the equation relating the total charge Q that flows through a search coil (Conceptual Example 29.3 in Section 29.2) to the magnetic field B. The search coil has N turns, each with area A and the flux through the coil is decreased from its initial value to zero in a time Dt. The resistance of the coil is R and the total charge is Q=I Dt where I is the average current induced by the change in flux

### From Example 29.3

• A search coil is a practical way to measure magnetic field strength. It uses a small, closely wound coil with N turns. The coil, of area A, is initially held so that its area vector A is aligned with a magnetic field with magnitude B. The coil is quickly pulled out of the field or rotated.

• Initially the flux through the field is F=NBA. When it leaves the field or rotated, F goes to zero. As F decreases, there is a momentary induced current which is measured. The amount of current is proportional to the field strength.

### Solution

• EMF=-DF/Dt

• DF=F2–F1

• Where F1 =NBA and F2 =0

• DF=-NBA

• EMF=NBA/Dt

• V=iR where V=EMF

• EMF=NBA/Dt=iR

• i=NBA/RDt

• Q=iDt=NBA/R

### Part B

• In a credit card reader, the magnetic strip on the back of the card is “swiped” past a coil within the reader. Explain using the ideas of the search coil how the reader can decode information stored in the pattern of magnetization in the strip.

### Solution

• The card reader contains a search coil.

• The search coil produces high EMF and low EMF as the card is swiped.

• A high EMF is treated as a binary 1 and a low EMF is treated as a binary 0

• Ascii information (character codes between 1 and 64) can be stored there in a few bits (6-bits or 26).

### Problem

A circular loop of flexible iron wire has an initial circumference of 165 cm but its circumference is decreasing at a rate of 12 cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with a magnitude of 0.5 T

• Find the EMF induced in the loop at the instant when 9 s have passed.

• Find the direction of the current in the loop as viewed looking in the direction of the magnetic field.

### EMF=-DF/Dt=-B*DA/Dt

• F=BA

• dF/dt=BdA/dt

• c=2pr so A=pr2

• r=c/2p so A=c2/4p

• dA/dt=d/dt(c2/4p)= (2c/4p)dc/dt

• dF/dt= B(c/2p)dc/dt

• Where B=0.5

• dc/dt=0.12 m/s

• At 9s, c=1.65-.12*(9s)=0.57 m

• dF/dt=5.44 x 10-3 V

• Since F is decreasing, the EMF is positive. Since EMF positive, point thumb along A and look at fingers.

• They curl counterclockwise.

### Problem

Suppose the loop in the figure below is

• Rotated about an edge parallel to the z-axis.

What is the maximum induced EMF in each case if A=600 cm2, w=35 rad/s, and B=.45T?

### Case b—rotating about the x-axis

• The normal to the surface is in the same direction as B during this rotation. Thus BA=constant

• d/dt(constant)=0 so EMF=0

### Problem

The figure below shows two parallel loops of wire having a common axis. The smaller loop (radius r) is above the larger loop (radius R) by a distance x>>R. Consequently, the magnetic field due to current i in the larger loop is nearly constant throughout the smaller loop. Suppose that x is increasing at a constant rate, v.

• Determine the magnetic flux though area of the smaller loop as a function of x.

• In the smaller loop find

• The induced EMF

• The direction of the induced current.

x

i

### Magnetic Field of a Circular Loop

By the RH Rule, the field is upward in the small ring

### Direction of current

• Assume that normal to smaller loop is positive upwards

• B, then, is in positive direction

• But F is decreasing or has a negative d dF /dt

• A negative * negative=positive so EMF is +

• RH Thumb in up direction, fingers curl counter clockwise!