Solutions

1

(a) We know that the magnitude of the force on a moving charged particle is given as F = qvB sin(Q_v,B). For this problem, we get

F = (1.6 x 10^-19 C)(3.0 x 10⁶ m/s)(0.3 T) sin (37^o) = 8.7 x 10^-14 N.

(b) Newton's second law tells us that the acceleration of the proton is given by

a = F/m = 8.7 x 10^-14 N/1.7 x 10^-27 kg = 5.1 x 10¹³ m/s².

2

Near the equator, the Earth has a magnetic field that is approximately 1 gauss = .0001 T in the northerly direction. Therefore, a wire with a current running through it that is aligned with the equator will experience a force on it that has a magnitude of

F = B I l where

I = current in the wire, and
l = length of the wire

since the angle between the magnetic field and the current is 90^o. In order to levitate this wire, this force must be equal in magnitude to the weight of the wire, i.e.

F = mg = (l) l g where l = linear mass density.

Setting these two forces equal to one another, we get B I l = (l) l g. This gives

I = (l) g/B = (.0010 kg/m)(9.8 m/s²)/(.0001 T) = 98 A.

Since we need the magnetic force to point upward, we get that the current should flow in an easterly manner from the righthand rule of thumb.

3

We know that the maximum magnetic force (and therefore, the minimum force required) occurs when the current and the magnetic field are perpendicular to one another. When this is true, the force on the wire will be F = B I l. From classical mechanics, we know that the frictional force on the wire will be

F_k = m_k m g = m_kl l g.

Setting these two forces equal to one another, we get

B I l = m_kl l g.

Solving for the magnitude of the magnetic force, we get

B = m_kl g/I = (.20)(.1 kg/m)(9.8 m/s²)/(1.50 A) = .13 T.

Since the current is flowing eastward, and the motion of the wire is northward, the magnetic field must be downward by the righthand rule of thumb.

4

The first thing to realize about this problem is that we are looking for the time that it takes the proton to make one circular orbit, i.e. 2pr/v. If the proton is moving perpendicularly to the magnetic field, then the force on the proton is F = q v B. Since the proton is moving in a circular orbit, this force must equal the centripetal force, i.e. mv²/r = qvB. Solving this for r/v, we get

r/v = m/(qB) = 1.7 x 10^-27 kg/((1.6 x 10^-19 C)(.758 T)) = 1.4 x 10^-8 s.

Multiplying this by 2 p, we get that the time that it takes to complete an orbit is 8.8 x 10^-8 s.

5

Due to the symmetry of the problem, we know that the magnetic field in and outside of the cable runs in circles around the center of the cable. Furthermore, symmetry also tells us the magnetic field on those circles is constant. Therefore, we can use Ampere's law in the following way to find the fields at A and B: Draw a circle through the point that is concentric with the cable. On those circles, the magnetic field is constant. Ampere's law states that

SB_|| (Dl) = B_|| (sum of Dl) = B_|| (2 pr) = m_o I_contained.

For A, this means

B (6.28)(.001 m) = m_o(1 A).

Solving for B, we get

B_A = (2 x 10^-7 Tm/A)(1 A)/(.001 m) = 2 x 10^-4 T.

The direction of the field is counterclockwise (righthand rule of thumb). At B, we change the radius to r = .003 m and the current to 2 A (3A into the page, 1 A out). Thus, we get

B_B = (2 x 10^-7 Tm/A)(2 A)/.003 m) = 1.3 x 10^-4 T.

The direction is clockwise.

If you find an error in these, or just have a comment, send an e-mail message to jpratte@kennesaw.edu