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Physics 108

Answers - Magnetic Fields

1.

Fm(N) = q (C) v(m/s) B sin symbol for anglev, B.
The unit of B is then N/(C-m/s)  or  N/(C/s-m) = N/A-m.
So 1 T = 1 N/(C/s-m) = 1 N/A-m.

2.

With Fm = q(v x B), Fm is always perpendicular to v and to B and to the plane that contains v and B, but the angle between v, B (symbol for anglev, B) can have any value.

3.

 





Let the magnetic field B be out of the page, south be down and north be up in the plane of the page so that v is up. The symbol for anglev, B = 900.

  1. Fm = qvB sin 90o = 1.6 x 10-19 C(3.2 x 107 m/s)(1.2 x 10-3 N/C/s-m)(1) = 6.1 x 10-15 N.

  2. If you point the fingers of your right hand up (in the direction of v) and then curl them toward B (out of the page), your thumb points to the right or to the east.


4.



The magnetic force Fm is in the same direction as the acceleration a or along the X-axis (See Fig. for #4 above).  Fm is perpendicular to both the velocity v and the magnetic field B, so B must lie in the Y-Z plane.  Fm = qvB sin symbol for anglev, B.  For the minimum B to produce a given force,  symbol for anglev,B = Q  in Fig. for #4  must be 90o.  Thus B must be along the Y-axis and it must be in the direction of -Y to produce a force and acceleration in the +X-direction when the velocity is in the
+ Z-direction.

Fm = ma = Fm = 1.67 x 10-27 kg(2 x 1013 m/s2) = 3.34 x 10-14 N
= qvB = (1.6 x 10-19 C)(107 m/s)B.   B = 0.021 T.

5.
  1. Fm = qvB sin symbol for anglev,B.  The maximum magnitude of force occurs when the angle between v and B is 90o or 270o; the minimum force is zero when the angle equals 0o or 180o.

    Fm = qvB sin 90o = 1.6 x 10-19 C(2 x 107 m/s)(0.080 N-s/C-m)(1)
         = 2.56 x 10-13 N.

  2. Fm= ma = 9.1 x 10-31 kg(1.41 x 1017 m/s2) = 12.8 x 10-14 N =
    qvB sin symbol for anglev,B = (2.56 x 10-13 N)sin symbol for anglev,B
     sin symbol for anglev,B = 12.8 x10-14 N/(2.56 x 10-13 N) = 0.50.
    symbol for anglev,B = 30o.

 

6.

   Fm = qvB sin symbol for anglev,B = (10-6 C)(103 m/s)(2 N-s/C-m)(0.50) = 10-3 N.

7.



  1. In Fig. for #7a above, with the velocity v parallel to the magnetic field B,
    symbol for anglev,B = 0 and sin 0o = 0, so there is no force on the electron. The electron continues to move with velocity v into the page.

  2. In Fig. for #7b above, the velocity v is perpendicular to the magnetic field B,  symbol for anglev,B = 90o and sin 90o = 1.  At the instant shown in the figure, the magnetic force Fm is down.

    Note: the force on a negative charge is opposite to that on a positive charge. Since this force is perpendicular to the velocity, it does not change the magnitude of the velocity, but it changes its direction.  As the direction of the velocity changes, the direction of  Fm changes so it always remains perpendicular to the velocity and the path is a circle.

  3. At some other angle with the magnetic field, the electron will continue to move with constant speed in the direction of the component of B which is parallel to velocity of the electron and proceed in a circle from the force due to the component of B which is perpendicular to the velocity of the electron. The resultant path is a spiral.

  4. There is an acceleration of the electron in parts (b) and (c).

8.
Fnet = ma
qvB sin 90o = mv2/r  or  v = qBr/m
vA = qBrA/m  and  vB = qBrB/m
v = 2pr/T, where T is the period of the motion.
vA = 2prA/TA = qBrA/m.   TA = 2pm/qB
vB = 2prB/TB = qBrB/m.   TB = 2pm/qB = TA.

9.

From #8,  r = mv/qB.  r2 = m2v/qB.  r1 = m1v/qB.  m2/m1 = r2/r1 = 22/20.

10.



  1. Fe = kQq/r2 = 9.0 x109 N-m2/C2(10-4 x 2 x 10-5)C2/104m2 =
    18 x 10-4 N  from q to - Q.

  2. Fm = qvB sin 90o = 2 x 10-5(4)N-s/m v

  3. If the particle moves

    1. clockwise the magnetic force is away from the center of the circle
      (Fig. for #10b)  and

    2. counterclockwise the magnetic force is toward the center of the circle (Fig. for #10a).

  4. Fnet = ma
    Using Fm into the center and positive as in Fig. for #10a,

    kQq/r2 + qvB = mv2/r        or
    mv2/r - qBv - kQq/r2 = 0.

    v = {(qB) ± [(qB)2 + 4mkqQ/r3]1/2}/2m/r

    v =

    v = {8 ± [784]1/2}m/s/2
       = {8 ± 28}m/s/2 = 18 m/s   or  -10 m/s.

    The velocity of 18 m/s corresponds to counterclockwise motion of the particle and the negative velocity of 10 m/s to clockwise motion of the particle.

11.
  1. As the positive particle with charge +q, mass m, and velocity vo down enters Region 1 with an electric field E to the right (the positively charged particle would be urged from the positive plate toward the negative plate), it experiences an electric force Fe to the right equal to qE. In order for the +q to pass through the electric field with no deflection, there must be a magnetic force Fm to the left. This magnetic force must be perpendicular to the magnetic field B and to the velocity vo.

    For vo down and Fm to the left,  B cannot be up or down or to the right or left because B is always perpendicular to Fm and in this problem B is said to be perpendicular to v.  B must be out of the page because only then will rotating v into B give Fm to the left.  Fm = q(vo x B).  Since we are told that B is perpendicular to vo,  symbol for anglevo, B = 90o and sin 90o = 1.  Thus,  Fm = qvoB. For no deflection, Fnet = ma = 0 because a = 0. Taking to the right to be positive, qE - qvoB = 0  or  B = E/vo = 106 N/C/(106 m/s) = 1.0 N-s/C-m = 1.0 N/A-m = 1.0 T out of the page.

  2. If,

    1. v = 5 x 105 m/s < 106 m/s,  qE > qvB and the particle is deflected to the right.

    2. v = 5 x 106 m/s > 106 m/s,  qvB > qE and the particle is deflected toward the left. This arrangement "selects" a velocity of a particle, which allows it to enter the slit of Region II.  For this reason, the apparatus is called a "velocity selector."

  3. As the particle enters region 2 without an electric field, it experiences a force to the right, which makes it go in a counterclockwise circle. For a velocity down and a magnetic force to the right, the new magnetic field B’ must be into the page. Again B’ is perpendicular to vo and the magnetic force which produces a centripetal acceleration is Fm = qvB’.

    Now,
       Fnet = ma, where a is the centripetal acceleration
    qvoB’ = mvo2/R   or
    R = mvo/qB’ = 2 x 10-12 kg(106 m/s)/(10-6 C)(2 N-s/C-m) = 1 m.


 

12.

 The magnetic force Fm is always perpendicular to the plane that contains the velocity v and the magnetic field B. In Fig. for #12a below,  F1 is out of the page and v1 is in the plane of the page. Thus B must be in the plane of the page. In Fig. for #12b below,  v2 is into the paper and F2 is in the plane of the paper. The plane of v2 and B must be perpendicular to the plane of the paper and F2. The direction of B is shown in Fig. for #12b.


  1. For Fig. a,  symbol for anglev1,B = 37o.
    F1 = qv1B sin 37o = 3 x 10-3 N = 10-6 C(500 m/s)(B)(3/5).
    B = 10 N-s/C-m.

  2. Now symbol for anglev2,B = 90o.
    F2 = qv2B sin 90o = 10-6 C(500 m/s)(10 N-s/C-m)(1) = 5 x 10-3 N.

  3. symbol for anglev3,B = 53o.  F3 is into the page.
    F3 = qv3B sin 53o = 10-6 C(103 m/s)(10 N-s/C-m)(4/5) = 8 x 10-3 N.

  4. F2 = qvB = mv22/r.
    r = mv2/qB = 10-8 kg(500 m/s)/(10-6 C)(10 N-s/C-m) = 0.5 m.
    It moves in a plane perpendicular to the plane of the page, which makes an angle of 53o with v1.

13.



  1. The Fig. for #13 above shows a side view of the inclined plane and the bar that carries a current out of the page. The forces acting on the bar are the magnetic force Fm to the left, the gravitational force mg, and the normal force N.

  2. Fm = ILB.  For the bar to remain at rest,  (Fnet)y = 0  and
    (Fnet)x = mg sin 22.5o - Fm cos 22.5o = 0  or  mg tan 22.5o = ILB.
    I = mg tan 22.5o/LB = (1.2 kg)(9.8 m/s2)(0.414)/(2.0 m)(0.50 N/A-m)
    = 4.87 A.

14.



I have taken up as north in Fig. for #14. In b of this figure, you see that the compass points north. When the compass needle is placed above a long current carrying wire (Fig. for #14c) the needle points to the west. Point the thumb of your right hand down. Above the wire your fingers and the magnetic field lines are counterclockwise. The tangent to the field lines is to the west, the direction of the magnetic field. The current in the wire that produces this deflection of the compass needle is south.

15.
The magnetic field lines due to a current I into the page are clockwise. The tangent to the magnetic field line at any point gives the direction of the field at that point.

At  P1,  the magnetic field is down.
At  P2,  the magnetic field is to the left.

16.

 The magnetic field B due to a very long wire is directly proportional to the current  I and inversely proportional to the distance r from the wire.
  1. If I is doubled, B is doubled.  If r is halved, B is also doubled.  
    If I is doubled and r is halved,  B goes up by a factor of 4.
    The new B is 4 x 0.50 T = 2.0 T.

  2. If I is halved, B is halved.  If r is doubled, B is halved.
    If I is halved and r is doubled,  B goes down by a factor of 4.
    The new B is 0.50 T/4 = 0.125 T.

17.


  1. The magnetic field lines for Ia (labeled a) and the magnetic field lines for Ib (labeled b) are shown in the Fig. for #17 above.  Ia = Ib = I.  At P1, the magnetic field due to Ia is Ba = µoI/2pr = Bb, the magnetic field due to Ib. Bb = - Ba.  The resultant magnetic field at P1 is zero.

  2. At P2,  the magnetic field Ba due to Ia equals µoI/2pr, and the magnetic field Bb due to Ib equals µoI/2p (3r), where r is the distance of P2 from Ia. Both fields are up.  The total field at P2 = µoI/2pr + µoI/6pr = 2µoI/3pr up.

  3. A proton out of the page at P2 experiences a force to the left since v is out of the page and B is up.

18.



The magnetic field lines for very long current-carrying wires are circles with the wire at the center. To find the direction of the field line, point the thumb of your right hand in the sense of the current and your fingers curl in the direction of the field.

In the figure for both I1 and I2 the direction of the field lines are counterclockwise. Notice I have drawn both so they pass through P because that is where I wish to find the field. The magnetic field at P is tangent to the circle at P.

For I1,  the circle has AP as a radius and the field due to it B1 is perpendicular to the radius and tangent to the field line so it is to the left and up, as shown in the figure above.

Similarly, the radius for the field lines for I2 is CP, and B2 is perpendicular to CP and tangent to its field line, so it is to the left and down.

The angle between B1 and B2 is 60o because B1 is perpendicular to AP and B2 is perpendicular to CP. Two angles that have their sides mutually perpendicular are equal.

Since  I1 = I2 = 3 A  and  r = 1.0 m,  B1 = B2
= µoI/2pr = 2 x 10-7 T-m/A (3A)/1m = 6 x 10-7 T.

B1y = B2y = B1 sin 30o,  but B1y is positive, while B2y is negative so there is no component of B in the Y-direction.

B = 2B1x = 2B2x = 2B1 cos 30o = 2B2 cos 30o =
2(6 x 10-7 T)(0.866) =10.4 x 10-7 T in the negative X-direction.

19.




In Fig. for #19a above,  I1 sets up a field B1 at P1 up.  A magnetic field line with I1 at the center would be counterclockwise.  A tangent to this field line at P1 is up. When I2 is placed at P1,  it experiences a force to the left.

In Fig. for #19b above,  I2 sets up a field B2 at P2 down.  A magnetic field line with I2 at the center would be counterclockwise.  A tangent to this field line at P2 is down. When I1 is placed at P2,  it experiences a force to the right.

The field approach correctly predicts that two wires carrying currents in the same direction attract each other.


20.



The magnetic field due to a very long wire is B = µoI/2pr.

For the long wire carrying a current I = 10 A at the position of side ea of the loop with r = 0.01 m = 10-2 m,
B = 2 x 10-7 N/A2(10A)/10-2 m = 2 x 10-4 N/A-m.

At the position of the side of the loop cd, with r = 0.03 m = 3 x 10-2 m,
B = 2 x 10-7N/A2(10A)/(3x10-2m) = 2/3(10-4)N/A-m.

Along ed and ac, the field varies with the distance r from the wire, but is the same for the same r.

To the right of the long wire, the magnetic field of the wire is into the plane of the paper.

The force Fm on a wire carrying a current I’ with length L in a magnetic field B is Fm = I’(L x B), where the direction of L is taken in the sense of I’ in that wire. For all lengths of the wire,  symbol for angleL, Bdue to the long wire  is 90o and sin 90o = 1. For the same value of r,  B at that position on the length de or ac is the same. The force on an element of length there is the same magnitude, but opposite in direction because Lde is to the left and Lac to the right. The net force on Lde and Lac is zero.

Fea = (20 A)(3 x 10-2 m)(2 x 10-4 N/A-m) sin 90o = 12 x 10-5 N to the right.

Fcd = (20 A)(3 x 10-2 m)(2/3 x 10-4 N/A-m) sin 90o = 4 x 10-5 N to the left.

Net force on loop = (12 - 4) x 10-5 N = 8 x 10-5 N to the right.

21.



There will be equal forces up and down on the upper and lower parts of the loop, respectively, but they will not produce any torque about the Y-axis. The force on length cd with I = 1.0 A, L = -0.25mj, and B = 0.5 N/A-m i is
Fm = 1.0 A(-0.25 m j x 0.5 N/A-m i) = 0.125 N k.  The torque t  = (r x F), where r is drawn from the axis to the point of application of the force F. The angle between r and Fm is 60o. The magnitude of r is 0.20 m and the magnitude of Fm is 0.125 N.  t = (0.20 m)(0.125 N) sin 60o = 0.022 N-m.  The direction of the torque from  t  = r x Fm  is in the negative y or - j direction.

22.



To find the direction of the field at P, point the thumb of your right hand at the upper (u) portion of the loop, out of the page. For this element, you can think of the magnetic field lines as being a concentric circle about it.  At P, the tangent to the line would be in the direction of Bu. Now go to an element at a lower portion of the loop. Again point your thumb there in the sense of the current or into the page.  At P, the field due to this current at P is shown as Bl. For every other element of the loop, the magnetic field vectors will make a cone about P. The components of the vectors in the Y and Z direction cancel, leaving the direction of the field in the positive X direction. When you look at a circular coil carrying a current in a counterclockwise direction, the magnetic field is along the axis of the coil, outward.

23.

 If the velocity v is parallel to the axis of the loop,  it is parallel to the magnetic field,  symbol for anglev,B = 0, and the force on it is zero. The electron will continue merrily on its path.

24.



  1. For a very long solenoid, the magnetic field inside the solenoid, far from either end, is a constant with Bsolenoid = µonIsolenoid, where n is the number of turns per unit length.

    B = (4p x 10-7 N/A2)(1000/m)(0.20 A) = 2.5 x 10-4 N/A-m = 2.5 x 10-4 T.

    When viewed from the right end of the solenoid, the current is counterclockwise. The magnetic field points to the right, as shown in the figure above.

  2. The magnetic field due to the very long wire Bwire = µoIwire/2pr.
    For a current in the wire to the left, below the axis of the solenoid at P the magnetic field due to the current-carrying wire is out of the page. The resultant field due to both makes an angle of 45o when
    Bsolenoid =  Bwire   or
    µonIsolenoid =  µoIwire/2pr   or
    r = (Iwire/Isolenoid)/2pn
      = (6.0/0.2)/(2000p m-1)
      = 4.8 x 10-3 m = 4.8 mm.

25.



In the figure above the dashed circles represents paths around which you can evaluate Ampere's law. There is also a path for a < r < b that is solid with arrows to denote a magnetic field. Notice that symbol for angleB,ds = 0.

From Ampere's law,  (current enclosed)


By symmetry for all cases,  symbol for angleB,ds = 0, and B is constant everywhere on the surface so,  


In general,  B2pr = µo (current enclosed)
  1. For r ≤ a,  B2pr = µo (0) = 0

  2. For a ≤ r ≤ b,  the current enclosed = NI, where N = the number of turns. B2pr = µo NI  and  B = µoNI/2pr.

  3. For r ≥ b, the current enclosed = I - I = 0  and  again B = 0.

26.

 There are two problems concerning magnetic fields, just as there were for electric fields.
  1. Given a distribution of current elements, find the magnetic field due to these current elements. There are three approaches to this problem:

    1. use the field due to a single long wire and then find the field due to a number of them by vector addition;

    2. use the Biot-Savart law for unsymmetrical situations;  and

    3. use Ampere's Law for symmetrical situations.

  2. Given a magnetic field, find the force on

    1. a moving charged particle, Fm = q(v x B)
      for a positively charged particle  and

    2. a current-carrying wire of length L,  Fm = I(L x B).

27.
  1. The force dF = (6 A)(dx)B sin 90o = (6 A)(dx m)(1 + 3x2 m-2)N/A-m(1).

  2. Unit of dF is N. The direction of dF is in the +Z-direction.

  3. dty = x dF sin 90o = x(m) (6 A)(dx m)(1 + 3x2 m-2)N/A-m
          = 6x(1 + 3x2 m-2)N-m.

  4. dtz = 0.  dF is parallel to the Z-axis.

  5. F = 6N / m(1 + 3x2m-2)dx = 6(0.5 + 0.125)N = 3.75N.

28.



The Biot-Savart Law states that ,
where  d symbol for scripted el bolded is in the direction of the sense of the current I.

The r with the inverted triangle over it is called r hat.  It is a unit vector drawn from  d symbol for scripted el bolded toward the point P where we wish to find the field.

For Fig. for #28 above, the segments AC and DE do not contribute to the field at P because the angle between  symbol for scripted el bolded and r hat is 0o for AC and 180o for DE and sin 0o = sin 180o = 0.

For the arc of length d symbol for scripted el bolded subtended by dQ,  the angle between d symbol for scripted el bolded and r hat is 90o and sin 90o = 1.    |d symbol for scripted el bolded x symbol for r hat| = d 

The definition of an angle in radians, dQ= d symbol for scripted el/R  or  dsymbol for scripted el = RdQ.
Thus dB = µo I R dQ/4pR2 = µoI dQ/4pR.

For an angle of Q = 150o = 5p/6 rad for the entire arc CD,
B = 5µoIp/24pR = 5µoI/24R.
Direction of B is the same as  d symbol for scripted el bolded x symbol for r hat,  or into the page.

29.

 From the Biot-Savart Law,  




For the circular loop,  symbol for angled symbol for scripted el bolded,  r hat = 90o and sin 90o = 1.
d symbol for scripted el bolded x  r hat  =  d symbol for scripted el  out of the page.

Since this is true for any element along the circular loop, the resultant field due to the semicircle is out of the page.  Also the distance r from each element of the semicircle to P is R.

B = µo I/4pR2 ∫ ds = µoI/4pR2 (pR) = µo I/4R,  since the length of the semicircle is pR.

We have found that the field of an infinite wire a distance r = R from it is given by B = µoI/2pR. The field due to a semi-infinite wire is µo I/4p R.  Both the upper and lower wires produce a field at P out of the page. Point the thumb of your right hand in the direction of the sense of the current in either the upper and lower wire and at P your fingers will point out of the page.

The total field due to both is  µo I/2p R.
The resultant field due to the upper and lower wires and the semicircle is
Btotal = (µo I/2R)[1/p + 1/2).

30.



For  r ≤ a,  the current density = Jr≤a = I/pa2
 Current enclosed by circle of radius r = (l/pa2)pr2 = Ir2/a2.

For  a ≤ r ≤ b,  current enclosed = I

For  b ≤ r ≤ c,  the current density = Jb≤r≤c = - I/p(c2-b2)
Current enclosed = I[1 - p(r2 - b2)/p(c2 - b2)] = I[(c2 - r2)/(c2 - b2)].

For  r ≥ c,  current enclosed = I - I = 0.

Only the cross-sections of the coaxial cylinders are shown in the figure above. The dashed circles in the figure are the paths drawn to evaluate
= µoI.

By symmetry, B is tangent to the path and constant for a given r.
For  r ≤ a  or  a ≤ r ≤ b,  the magnetic field lines are counterclockwise for the current of the inner cylinder out of the page.  B, dsymbol for scripted el bolded= 0.
Since B is constant over the path,    B (2pr).
This is true for all values of r.
  1. For r ≤ a,                   B2pr = µo(current enclosed) = µo(Ir2/a2),  and
                                          B = µo(Ir/2pa2)

  2. For a ≤ r ≤ b,             B2pr = µoI  and  B = µoI/2pr

  3. For b ≤ r ≤ c               B2pr = µoI[(c2 - r2)/(c2 - b2)]
                                          B = µoI[(c2 - r2)/(c2 - b2)]/2pr

  4. For r ≥ c                     B2pr = 0 and B = 0.

31.


Number of wires/length = N/w.  Number of wires in length dx = (N/w)dx.
  1. For length dx, dB = NµoI dx/2prw = NµoI dx/2pw(x2 + a2)1/2.

  2. If you point the thumb of your right hand in the direction of the current at x, the magnetic field line through point P is counterclockwise. The tangent to the field line at P is perpendicular to the radius r of the circle in the direction of dB in the figure above.

    For a corresponding dx on the left hand of the origin, the dB would be perpendicular to the line from dx to P. You can see from the figure that the components of the dB's in the Y-direction dBy cancel. The resultant component is in the negative X-direction.

    dBx = dB cos Q = Nµo(I/w) dx/2p(x2 + a2)1/2 (a/(x2 + a2)1/2)
                            = Nµo(I/w)a dx/2p(x2 + a2).

    Let x = a tan Q.
    dx = a sec2QdQ.
    dBx = Nµo(I/w)a2 sec2 QdQ/2p(a2 tan2 Q + a2)
           = Nµo(I/w)a2 sec2 QdQ/2pa2sec2 Q
           = (NµoI/2wp) dQ.

    The lower limit on x is - w/2 and the upper limit on x  is + w/2.
    The lower limit on Q is tan -1 (-w/2)/a  and the upper limit on Q
     is tan-1 (+ w/2)/a.

    B =(NµoI/2p) d Q
        = (NµoI/2wp)[tan-1 (+w/2)/a - tan-1 (-w/2)/a]
        = (NµoI/wp)[tan-1 w/2a].

32.
  1. The magnetic moment  µ = IpR2.

  2. B = µoIpR2/2p(R2 + z2)3/2  becomes
    B = µoµ/2p(R2 + z2)3/2.

  3. For z >> R,  you can neglect R in the denominator of B and then you have 2p(z2)3/2 = 2pz3 and B becomes
               m o m/2pz3.

  4. B = 2 x 10-7 N/A2 (9.27 x 10-24 A-m2)/(10-10 m)3 = 1.9 N/A-m = 1.9 T.

  5. The spinning electron gives a field and magnetic moment opposite in direction to a positive charge moving in a loop. Since both change sign, the net effect for the potential energy will be the same as for current loops. That is U = - m . B = - µB cos symbol for anglem,B.

    1. When they are aligned they will have minimum energy.

    2. Antialigned  U = µB
                             = 9.27 x 10-24 A-m2 x 1.9 N/A-m/(1.6 x 10-19 J/eV)
                             = 1.1 x 10-4 eV.

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Susan D. Kunk
Phyllis J. Fleming
August 8, 2002
May 1, 2003