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Solutions to Homework Questions 6

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Physics 112 <strong>Homework</strong> 6 (solutions) (2004 Fall)<br />

Chapt20, Problem-17: The person in Figure P20.17 is girded<br />

about the chest with a breathing moni<strong>to</strong>r, which is a 100-turn coil. As a breath<br />

is inhaled, the area of the coil varies from 0.120 m 2 <strong>to</strong> 0.124 m 2 . Earth’s<br />

magnetic field is 50.0 _T and makes an angle of 22.5° with respect <strong>to</strong> the<br />

normal <strong>to</strong> the loop. If the patient inhales a breath in 1.59 s, what is the<br />

average voltage induced in the coil during the inhalation<br />

Solution:<br />

( )<br />

! = N "# B NB ("A)cos$<br />

=<br />

"t "t<br />

= 100 ( 50.0 ! 10-6 T) [( 0.124 " 0.120) m 2<br />

]cos 22.5°<br />

1.59 s<br />

= 1.16 !10 "5 V= 11.6 µV<br />

Chapt20, Problem-19: A Boeing-747 jet with a wing span of 60.0 m is flying horizontally at a speed of<br />

300 m/s over Phoenix, Arizona, at a location where Earth’s magnetic field is 50.0 µT at 58.0° below the horizontal. What<br />

voltage is generated between the wingtips<br />

Solution:<br />

! = B " l v , where B ! is the component of the magnetic field perpendicular <strong>to</strong> the velocity v. Thus,<br />

( ) 300 m s<br />

[ ] 60.0 m<br />

! = ( 50.0 " 10 #6 T)sin 58.0°<br />

( )= 0.763 V<br />

Chapt20, Problem-22: A helicopter has blades of length 3.0 m, rotating at 2.0 rev/s about a central hub. If<br />

the vertical component of Earth’s magnetic field is 5.0x10 –5 T, what is the emf induced between the blade tip and the central<br />

hub<br />

Solution:<br />

During each revolution, one of the ro<strong>to</strong>r blades sweeps out a horizontal circular area of radius l, A = ! l 2 .<br />

The number of magnetic field lines cut per revolution is !" B = B # A = B vertical A. The induced emf is then<br />

! = "# B<br />

"t<br />

( )<br />

= B vertical<br />

$ l 2<br />

1 f<br />

( ) $ 3.0 m<br />

= 5.0 %10 &5 T<br />

0.50 s<br />

[ ( ) 2<br />

]<br />

= 2.8 % 10 &3 V =<br />

2.8 mV<br />

Chapt20, Problem-23: A bar magnet is positioned near a coil<br />

of wire as shown in Figure P20.23. What is the direction of the current through<br />

the resis<strong>to</strong>r when the magnet is moved (a) <strong>to</strong> the left (b) <strong>to</strong> the right<br />

Solution:<br />

(a) To oppose the motion of the magnet, the magnetic field generated<br />

by the induced current should be directed <strong>to</strong> the right along the axis of<br />

the coil. The current must then be left <strong>to</strong> right through the resis<strong>to</strong>r.<br />

(b) The magnetic field produced by the current should be directed <strong>to</strong><br />

the left along the axis of the coil, so the current must be right <strong>to</strong> left<br />

through the resis<strong>to</strong>r.<br />

2

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