Problem 1:
A copper wire (r = 1.72´10-8 Wm) has a length of 160 m and a diameter of 1.00 mm. If the wire is connected to a 1.5V battery, how much current flows through the wire?
Solution:
 | Concepts, principles, relations that apply to the
problem: Ohm's law, resistance of a wire |
| Why do they apply? The current can be found from Ohm's Law, V = IR. V is the battery voltage, so if R can be determined, the current can be calculated. The resistance of the wire is R = rl/A. |
| How do they apply? For copper r = 1.72´10-8 Wm. The cross-sectional area of the wire is A = pr2 = p(0.0005)2 = 7.85´10-7 m2. The resistance of the wire then is ((1.72´10-8)´160/(7.85´10-7)) W = 3.5 W. The current is I = V/R = (1.5/3.5) A = 0.428 A. |
| Details of the calculation: None |
Problem 2:
Find the equivalent resistance between the points A and B of the circuit shown in the figure below.
Solution:
| Concepts, principles, relations that apply to the
problem: Kirchhoff's rules | ||||||||
| Why do they apply? We can find the equivalent resistance using Kirchhoff's rules. The junction rule states that the sum of the currents entering a junction must equal the sum of the currents leaving that junction. The loop rule states that the sum of the potential differences around any closed circuit loop must be zero. Assume some direction for the current flowing in each part of the circuit.
We can use the junction rule as many times as it is possible to include in it a current that has not been used in a previous junction rule equation. The number of equations must be equal to the number of unknowns. | ||||||||
| How do they apply? Assume you connect a battery between A and B so that A is at some voltage V and B is at ground. A current I will start flowing through the circuit from A to B. V = IR, R = V/I. If you know I, then you know R. To find R for the circuit we need to know the currents flowing through the 6 resistors. Let I(1) denote the current flowing through the 1 ohm resistor, I(2) denote the current through the 2 ohm resistor, and so on. The total current leaving point A is denoted by I. We need to use Kirchhoff's rules to find 7 equations for the seven currents, and then use algebra to solve those seven equations simultaneously. (1) For the junction labeled J1 we have I(1) + I(2) + I(5) - I = 0. (2) For the junction labeled J2 we have I(2) - I(7) - I(9) = 0 (3) For the junction labeled J3 we have I(13) - I(7) - I(5) = 0 (4) For loop 1 we have V - 1*I(1) = 0 (5) For loop 2 we have 1*(I1) - 2*I(2) - 9*I(9) = 0 (6) For loop 3 we have 9*I(9) - 7*I(7) - 13*I(13) = 0 (7) For loop 4 we have 2*I(2) + 7*I(7) - 5*I(5) = 0 Let us now use equation 4 to eliminate I(1) from the other equations. I(1) = V. We now have six equations. (1) For the junction labeled J1 we have V + I(2) + I(5) - I = 0. (2) For the junction labeled J2 we have I(2) - I(7) - I(9) = 0 (3) For the junction labeled J3 we have I(13) - I(7) - I(5) = 0 (5) For loop 2 we have V - 2*I(2) - 9*I(9) = 0 (6) For loop 3 we have 9*I(9) - 7*I(7) - 13*I(13) = 0 (7) For loop 4 we have 2*I(2) + 7*I(7) - 5*I(5) = 0 Let us now use equation 2 to eliminate I(2) from the other equations. I(2) = I(7)+I(9). We now have five equations. (1) For the junction labeled J1 we have V + I(7) + I(9) + I(5) - I = 0. (3) For the junction labeled J3 we have I(13) - I(7) - I(5) = 0 (5) For loop 2 we have V - 2*I(7) - 11*I(9) = 0 (6) For loop 3 we have 9*I(9) - 7*I(7) - 13*I(13) = 0 (7) For loop 4 we have 9*I(7) + 2*I(9) - 5*I(5) = 0 Let us now use equation 3 to eliminate I(5) from the other equations. I(5) = I(13)-I(7). We now have 4 equations. (1) For the junction labeled J1 we have V + I(9) + I(13) - I = 0. (5) For loop 2 we have V - 2*I(7) - 11*I(9) = 0 (6) For loop 3 we have 9*I(9) - 7*I(7) - 13*I(13) = 0 (7) For loop 4 we have 14*I(7) + 2*I(9) - 5*I(13) = 0 Let us now use equation 5 to eliminate I(7) from the other equations. 2*I(7) = V-11*I(9). We now have three equations. (1) For the junction labeled J1 we have V + I(9) + I(13) - I = 0. (6) For loop 3 we have -7*V + 95*I(9) - 26*I(13) = 0 (7) For loop 4 we have 7*V - 75*I(9) - 5*I(13) = 0 Let us now use equation 7 to eliminate I(9) from the other equations. 75*I(9) = 7*V - 5*I(13). We now have 2 equations. (1) For the junction labeled J1 we have 82*V + 70*I(13) - 75*I = 0. (6) For loop 3 we have 140*V - 2425*I(13) = 0 Let us now use equation 6 to eliminate I(13) from the other equations. 485*I(13) = 28*V. We now have 1 equations. (1) For the junction labeled J1 we have 41730*V - 36375*I = 0. I = (41730/36375)*V = 1.147*V R = V/I = 0.87 ohms The resistance is 0.87 ohms. | ||||||||
| Details of the calculation: None |
Problem 3:
A wire having a mass per unit length of 0.5 g/cm carries a 2 A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?
Solution:
| Concepts, principles, relations that apply to the
problem: The magnetostatic force on a wire,  |
| Why do they apply? We are to determine the minimum magnetic field needed to produce a force F = -mg. |
| How do they apply? Orient your coordinate system so that the x-axis points towards the south and the z-axis points upward. We want the direction of the magnetic force F= IL´B on the wire segment to be upward and the magnitude to be equal to mg. To get the maximum F for the minimum B, B hat to point into the y-direction. Then F = ILB.
To lift the wire we need |
| Details of the calculation: None |
Problem 4:
A rectangular loop consists of N closely wrapped turns and has dimensions a = 0.4 m and b = 0.3 m. The loop is hinged along the y-axis and its plane makes an angle of 30o with the x-axis.
What is the magnitude of the torque exerted on the loop by a uniform magnetic
field B = 0.8 T directed along the x-axis when the current I is 1.2 A in the
direction shown?
What is the expected direction of rotation of the loop?
Solution:
| Concepts, principles, relations that apply to the
problem: The torque on a current loop in a magnetic field |
| Why do they apply? We are asked to find the torque on a current loop. |
| How do they apply? The torque on a dipole is t = m´B. The torque exerted on the loop is t = m´B. The magnetic moment is the magnetic moment per turn, times the number of turns, m = NabI. The direction of m makes an angle of 60o with the x-axis and an angle of 150o with the z-axis. The smallest angle between m and B is 60o.
The magnitude of the torque therefore is |
| Details of the calculation: None |
Problem 5:
(a) A circular loop of wire of radius R carries a
current I. Find the magnetic induction B on the axis of the loop,
as a function of the distance z from the center of the loop.
(b) Use the result to find B at points on the axis of a solenoid
of radius R and length L wound with n turns per unit
length.
(c) Use this result to find the self-inductance of a very long solenoid (L >> R).
Solution:
| Concepts, principles, relations that apply to the
problem: The Biot-Savart law, the principle of superposition, self inductance |
| Why do they apply? We find the field of a single loop using the Biot-Savart law. The principle of superposition then yields the field of many loops. |
| How do they apply? (a) From the Biot-Savart law we have dBz(r) = (m0/4p)[Idl´(r-r')/|r-r'|3]×k. On the z-axis r = zk. Symmetry dictates that B = Bzk on the z-axis.  .dBz = (m0/4p)[IR2df/(R2+z2)3/2].  .At the center of the loop Bz = m0I/(2R), and for z >> R Bz = m0IR2/(2z3) µ 1/z3. Details of the calculation: (b) Assume N loops are stacked up, from z' = -z to z' = z, n per unit length.  . . cosq1 = (z+z1)/(R2 + (z+z1)2)1/2, cosq2 = (z-z1)/(R2 + (z-z1)2)1/2. Bz = (m0In/2)(cosq1 - cosq2) For the solenoid n = N/L, z1 = L/2. (c) At the center of the solenoid B = Bzk. Bz = (m0In/2)(2z1/(R2 + z12)1/2) = (m0IN/2)((R2 + (L/2)2)-1/2). For L >> R then Bz » (m0In). The field is approximately uniform inside the solenoid and approximately zero outside the solenoid. F = L1I, F = Flux, L1 = self inductance. F = BpR2N = m0InpR2N = L1I, L1 = m0n2pR2L = m0n2V, V = volume of the solenoid. |
Problem 6:
Show that the magnitude of the
magnetic induction B at the center of a loop of wire carrying a
current I and shaped like a regular plane polygon of 2n sides, the
distance between parallel sides being 2a, is
using the Biot-Savart law. The n = 3
case is shown.
Solution:
| Concepts, principles, relations that apply to the
problem: The Biot-Savart law, the principle of superposition |
| Why do they apply? We find the field of a single segment using the Biot-Savart law. The principle of superposition then lets us add the fields of the different segments. |
| How do they apply? The field B at the center of the circle due to the current in the wire from -l to +l is  (SI units).Here r = 0, r' = xi + aj, dl' = dxi, |r-r'| = (x2 + a2)1/2. dl' ´ (r-r') = (a dx) k  .For a regular plane polygon of 2n sides we have 2n wire segments. For each segment 2a = 2p/(2n), a = p/(2n). Symmetry dictates that B(0) due to each segment is the same.  .Details of the calculation: None |
Problem 7:
An eccentric hole of radius a is bored parallel to the axis of a right circular cylinder of radius b (b > a). The two axes are at a distance d apart. A current of I amperes flows in the cylinder. What is the magnetic field at the center of the hole? Assume a uniform current density.
Solution:
| Concepts, principles, relations that apply to the
problem: Ampere's law, the principle of superposition |
| Why do they apply? A cylinder of radius a carrying a uniform current density -jk superimposed on a cylinder of radius b carrying a uniform current density jk with the center of the smaller cylinder at d presents an equivalent problem. |
| How do they apply? Use Ampere's law: B(d) = m0pd2j/(2pd) = m0dj/2. j = I/(p(b2-a2)), B(d) = m0Id/(2p(b2-a2)). The direction of B is the f-direction. |
| Details of the calculation: None |
Problem 8:
In practical magnetic structures, uniform magnetic fields are frequently necessary. The uniformity of the field produced by Helmholtz coils, or two co-axial loops which carry currents in the same direction, is one of their most important characteristics. Assume that the coils have a radius a, have axes on the x-axis, carry current I each, and are separated by a distance b.
(a) Find the magnetic field at a point P on the axis of the loops and
a distance x from the midpoint O.
(b) Expand the expression for the field in a power series, retaining terms to
order x2.
(c) What relationship must exist between a and b such that the
x2 terms vanish? What is the significance of this?
(d) Show that the field created by the coils to this order and under the
conditions established in part c is given by Bx = 8I/(53/2ae0c2).
Solution:
| Concepts, principles, relations that apply to the
problem: The Biot-Savart law, a series expansion |
| Why do they apply? The magnetic field on the axis of a current loop is found from the Biot-Savart law. Expanding the field near x = 0 in terms of x, we can find the condition for the second order term to vanish. The first order term is be zero due to the symmetry of the situation, so B(x) = B(0) to second order if the term proportional to x2 vanished. |
| How do they apply? (a) B = B(x)i on the x-axis. B(x) = (m0Ia2/2)[a2 + (d-x)2]-3/2 + (m0Ia2/2)[a2 + (d+x)2]-3/2. Here d = b/2. (b) B(x) = B(0) + x(¶B/¶x)|0 + (x2/2)(¶2B/¶x2)|0 + ... ¶B/¶x =(m0Ia2/2)[-(3/2)2(d+x)[a2 + (d+x)2]-5/2 + (3/2)2(d-x)[a2 + (d-x)2]-5/2, (¶B/¶x)|0 = 0. ¶2B/¶x2 = (3m0Ia2/2)[5(d+x)2[a2 + (d+x)2]-7/2 - [a2 + (d+x)2]-5/2 + 5(d-x)2[a2 + (d-x)2]-7/2 - [a2 +( d-x)2]-5/2]. (¶2B/¶x2)|0 = (3m0Ia2/2)[10d2[a2 + d2]-7/2 - 2[a2 + d2]-5/2] = (3m0Ia2/2)[(8d2-2a2)[a2 + d2]-7/2]. |
| Details of the calculation: (c) For the second order term to vanish we need 8d2-2a2 = 0, d = a/2, b = a. The field is then uniform to second order. (d) B(x) = B(0) = (m0Ia2)[a2 + d2]-3/2 = (m0I/a)[1 + 1/4]-3/2 = (m0I/a)(4/5)3/2 = 8I/(53/2ae0c2). |
Problem 9:
Consider an idealized ion beam of radius R and length much longer than R.
Show that an individual ion at the periphery of this beam is subject to the net outward force F = (2Iq/Rv)(1 - v2/c2) where I is the beam current, q is the charge of each ion, and v is the velocity of the ions. Assume that the charge and current densities have cylindrical symmetry.
Solution:
| Concepts, principles, relations that apply to the
problem: Ampere's law, Gauss' law, the Lorentz force |
| Why do they apply? Because of the cylindrical symmetry we can find the electric and magnetic fields from Gauss' law and Ampere's law respectively. We can then find the force F = q(E +v´B) on an individual particle with charge q. |
| How do they apply? In SI units: Gauss' law: At r = R we have 2pRE = I/(ve0), E =I/(2pe0Rv). [The charge per unit length in the beam is l = I/v.] Ampere's law: At r = R we have 2pRB = m0I = I/(e0c2), B = I/(2pRe0c2). If the current flows in the z-direction then E points into the r direction and B points into the f direction. The force on an ion of charge q moving with velocity vk at r = R is F = q(E + v´B) F = qI/(2pe0Rv) - qIv/(2pRe0c2) = qI/(2pe0Rv)(1 - v2/c2) = (1/(4pe0))(2Iq/Rv)(1 - v2/c2). The force points in the r direction. In Gaussian units: F = 2Iq/(Rv)(1 - v2/c2). |
| Details of the calculation: None |
Problem 10:
A flat insulating disk of radius R carries a uniform charge density.
The disk rotates with angular velocity w about an axis through its center and
perpendicular to its plane.
(a) Find B at a point on the axis at a distance z from the disk.
(b) Find the magnetic dipole moment of the disk.
Solution:
| Concepts, principles, relations that apply to the
problem: The Biot-Savart law, the principle of superposition |
| Why do they apply? We can consider the disk to be made up of a series of rotating rings. Each ring constitutes a current loop. |
| How do they apply? (a) Consider a current loop of radius r in the z = 0 plane. Let the current flow in the f-direction. The field on the z axis is B(z) = (m0Ir2/2)[r2 + z2]-3/2 k. A ring shaped portion of the disk of thickness dr constitutes a current loop with dI = srwdr. [The current density is j = rv. For a surface current we have k = sv, v = rw.] Let w = wk. The magnetic field on the z axis due to the ring shaped portion of the disk is dB(z) = (m0r3swdr/2)[r2 + z2]-3/2 k. B(z) = k(m0sw/2)òr3dr/[r2 + z2]3/2 . The integral is from 0 to R. B(z) = k(m0sw/2)[(R2 + 2z2)/[R2 + z2]1/2 - 2|z|] |
| Details of the calculation: (b) For each ring shaped portion of the disk dm = dI A k = srwdr pr2. For the entire disk: m = òdm = swpòr3dr = swpR4/4. |
