Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 26, Problem 66P

(a)

To determine

The magnetic field

(a)

Expert Solution
Check Mark

Answer to Problem 66P

  1.36 T

Explanation of Solution

Given:

Number density of electrons =n=8.47×1022 cm3=8.47×1022×106m3=8.47×1028m3

Current in the metal strip =I=20 A

Width of the metal strip =b=2 cm = 0.02 m

Thickness of the metal strip =t=0.1 cm = 0.001 m

Area of cross-section of the strip =A

Magnitude of charge on each electron =e=1.6×1019C

Drift speed of electrons =vd

Hall voltage =V=2 μV=2×106 V

Formula Used:

Area of cross-section of a rectangular shape strip is given as

  A=bt

Current is given as

  I=neAvd

Hall voltage is given as

  V=Bbvd

Calculation:

Area of cross-section of the strip is given as

  A=bt

Current flowing through the strip is given as

  I=neAvdI=ne(bt)vdvd=Inebt                                                     Eq-1

Hall voltage is given as

  V=Bbvdusing Eq-1V=Bb(I nebt)V=B(I net)V(net)=BIB=neVtI

Inserting the values, we get

  B=(8.47× 10 28)(1.6× 10 19)(2× 10 6)(0.001)20B=1.36 T

Conclusion:

The magnetic field comes out to be 1.36 T

(b)

To determine

The magnetic field

(b)

Expert Solution
Check Mark

Answer to Problem 66P

  3.56 T

Explanation of Solution

Given:

Number density of electrons =n=8.47×1022 cm3=8.47×1022×106m3=8.47×1028m3

Current in the metal strip =I=20 A

Width of the metal strip =b=2 cm = 0.02 m

Thickness of the metal strip =t=0.1 cm = 0.001 m

Area of cross-section of the strip =A

Magnitude of charge on each electron =e=1.6×1019C

Drift speed of electrons =vd

Hall voltage =V=5.25 μV=5.25×106 V

Formula Used:

Area of cross-section of a rectangular shape strip is given as

  A=bt

Current is given as

  i=neAvd

Hall voltage is given as

  V=Bbvd

Calculation:

Area of cross-section of the strip is given as

  A=bt

Current flowing through the strip is given as

  I=neAvdI=ne(bt)vdvd=Inebt                                                     Eq-1

Hall voltage is given as

  V=Bbvdusing Eq-1V=Bb(I nebt)V=B(I net)V(net)=BIB=neVtI

Inserting the values, we get

  B=(8.47× 10 28)(1.6× 10 19)(5.25× 10 6)(0.001)20B=3.56 T

Conclusion:

The magnetic field comes out to be 3.56 T

(c)

To determine

The magnetic field

(c)

Expert Solution
Check Mark

Answer to Problem 66P

  5.42 T

Explanation of Solution

Given:

Number density of electrons =n=8.47×1022 cm3=8.47×1022×106m3=8.47×1028m3

Current in the metal strip =I=20 A

Width of the metal strip =b=2 cm = 0.02 m

Thickness of the metal strip =t=0.1 cm = 0.001 m

Area of cross-section of the strip =A

Magnitude of charge on each electron =e=1.6×1019C

Drift speed of electrons =vd

Hall voltage =V=8 μV=8×106 V

Formula Used:

Area of cross-section of a rectangular shape strip is given as

  A=bt

Current is given as

  I=neAvd

Hall voltage is given as

  V=Bbvd

Calculation:

Area of cross-section of the strip is given as

  A=bt

Current flowing through the strip is given as

  I=neAvdI=ne(bt)vdvd=Inebt                                                     Eq-1

Hall voltage is given as

  V=Bbvdusing Eq-1V=Bb(I nebt)V=B(I net)V(net)=BIB=neVtI

Inserting the values, we get

  B=(8.47× 10 28)(1.6× 10 19)(8× 10 6)(0.001)20B=5.42 T

Conclusion:

The magnetic field comes out to be 5.42 T

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