Magnetism and the Electric Field

Chapter 29.  Magnetism and the Electric Field

Magnetic Fields

29-1.        The area of a rectangular loop is 200 cm2 and the plane of the loop makes an angle of 410 with a 0.28-T magnetic field.  What is the magnetic flux penetrating the loop?  

A = 200 cm2 = 0.0200 m2;     θ = 410;    B = 0.280 T[pic 1]

φ = BA sin θ = (0.280 T)(0.0200 m2) sin 410;      φ = 3.67 mWb

[pic 2]

29-2.        A coil of wire 30 cm in diameter is perpendicular to a 0.6-T magnetic field. If the coil turns so that it makes an angle of 600 with the field, what is the change in flux?

[pic 3];     A = 7.07 x 10-2 m2;     Δφ = φf - φo

[pic 4]

[pic 5][pic 6]

Δφ = φf - φo = 36.7 mWb – 42.4 mWb;      Δφ = -5.68 mWb

[pic 7]

29-3.        A constant horizontal field of 0.5 T pierces a rectangular loop 120 mm long and 70 mm wide.  Determine the magnetic flux through the loop when its plane makes the following angles with the B field: 00, 300, 600, and 900.    [Area = 0.12 m)(0.07 m) = 8.40 x 10-3 m2 ]

φ = BA sin θ;     BA = (0.5 T)(8.4 x 10-3 m2) = 4.2 x 10-3 T m2[pic 8][pic 9]

        φ1 = (4.2 x 10-3 T m2) sin 00 =   0 Wb;        φ2 = (4.2 x 10-3 T m2) sin 300 =  2.10 mWb;[pic 10][pic 11]

        φ3 = (4.2 x 10-3 T m2) sin 600 =  3.64 mWb;        φ1 = (4.2 x 10-3 T m2) sin 900 =  4.20 mWb

29-4.        A flux of 13.6 mWb penetrates a coil of wire 240 mm in diameter.  Find the magnitude of the magnetic flux density if the plane of the coil is perpendicular to the field.

[pic 12];     A = 4.52 x 10-2 m2;      φ = BA sin θ

[pic 13] ;     B = 0.300 T[pic 14]

[pic 15]

29-5.        A magnetic flux of 50 μWb passes through a perpendicular loop of wire having an area of 0.78 m2. What is the magnetic flux density?

[pic 16] ;          B = 64.1 μT[pic 17]

[pic 18]

29-6.        A rectangular loop 25 x 15 cm is oriented so that its plane makes an angle θ with a 0.6-T B field.  What is the angle θ if the magnetic flux linking the loop is 0.015 Wb?

A = (0.25 m)(0.15 m) = 0.0375 m2;      φ = 0.015 Wb

[pic 19];      θ = 41.80[pic 20]

The Force on  Moving Charge

29-7.        A proton (q = +1.6 x 10-19 C) is injected to the right into a B field of 0.4 T directed upward. If the velocity of the proton is 2 x 106 m/s, what are the magnitude and direction of the magnetic force on the proton?[pic 21]

        F = qvB⊥ = (1.6 x 10-19 C)(2 x 106 m/s)(0.4 T)[pic 22]

                        F = 1.28 x 10-13 N, into paper

29-8.        An alpha particle (+2e) is projected with a velocity of 3.6 x 106 m/s into a 0.12-T magnetic field.  What is the magnetic force on the charge at the instant its velocity is directed at an angle of 350 with the magnetic flux?   [ q = 2 (1.6 x 10-19 C) = 3.2 x 10-19 C ][pic 23]

F = qvB sinθ = (3.2 x 10-19 C)(3.6 x 106 m/s)(0.12 T) sin 350;    F = 7.93 x 10-14 N

[pic 24]

29-9.        An electron moves with a velocity of 5 x 105 m/s at an angle of 600 with an eastward B field.  The electron experiences an force of 3.2 x 10-18 N directed into the paper. What are the magnitude of B and the direction the velocity v?[pic 25]

        In order for the force to be INTO the paper for a NEGATIVE[pic 26]

        charge, the 600 angle must be S of E.     θ = 600 S of E

[pic 27];       B = 46.3 μT[pic 28]

[pic 29]

29-10.        A proton (+1e) is moving vertically upward  with a velocity of 4 x 106 m/s.  It passes through a 0.4-T magnetic field directed to the right. What are the magnitude and direction of the magnetic force?    [pic 30]

                [pic 31];    [pic 32]

                             F = 2.56 x 10-13 N, directed into paper.

[pic 33]

29-11.        What if an electron replaces the proton in Problem 29-10.  What is the magnitude and direction of the magnetic force?

        The direction of the magnetic force on an electron is opposite to that of the proton, but the magnitude of the force is unchanged since the magnitude of the charge is the same.[pic 34]

Fe = 2.56 x 10-13 N, out of paper.

*29-12. A particle having a charge q and a mass m is projected into a B field directed into the paper. If the particle has a velocity v, show that it will be deflected into a circular path of radius:

[pic 35]

Draw a diagram of the motion, assuming a positive charge entering the B field from left to right.  Hint: The magnetic force provides the necessary centripetal force for the circular motion.[pic 36]

              [pic 37]

                            From which:    [pic 38]

The diagram shows that the magnetic force is a centripetal force that acts toward the center causing the charge to move in a counterclockwise circle of radius R.

[pic 39]

*29-13.        A deuteron is a nuclear particle consisting of a proton and a neutron bound together by nuclear forces. The mass of a deuteron is 3.347 x 10-27 kg, and its charge is +1e.  A deuteron projected into a magnetic field of flux density 1.2 T is observed to travel in a circular path of radius 300 mm. What is the velocity of the deuteron?  See Problem 29-12.

[pic 40]

[pic 41];      v = 1.72 x 107 m/s[pic 42]

Note: This speed which is about 6% of the speed of light is still not fast enough to cause significant effects due to relativity (see Chapter 38.)

Force on a Current-Carrying Conductor

29-14.        A wire 1 m in length supports a current of 5.00 A and is perpendicular to a B field of 0.034 T. What is the magnetic force on the wire?[pic 43]

F = I B⊥ l= (5 A)(0.034 T)(1 m);      F = 0.170 N

[pic 44]

29-15.        A long wire carries a current of 6 A in a direction 350 north of an easterly 40-mT magnetic field.  What are the magnitude and direction of the force on each centimeter of wire?[pic 45]

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