Problem 1:
A solenoid is designed to store UL = 0.10 J of energy when it carries a current of I = 450 mA. The solenoid has a cross-sectional area of A = 5.0 cm2 and a length l = 0.20m. How many turns of wire must the solenoid have?
Solution:
 | Concepts, principles, relations
that apply to the problem: The magnetic field inside a solenoid, magnetostatic energy |
| Why do they apply? Approximation: B = m0nI inside the solenoid, B = 0 outside.  The integral over all space reduces to an integral over the volume of the solenoid. |
| How do they apply? UL = (1/(2m0))B2V = m0n2I2lA/2. 2UL/( m0I2lA) = n2, n = 88654. N = nl = 17731. or UL = ½LI2, L = nlBA/I = m0n2lA, n2 = 2UL/( m0I2lA). |
| Details of the calculation: None |
Problem 2:
Suppose that a very long coaxial line is divided into three
regions
(i) current I into the page for 0 < r < a,
(ii) current 0 for radius a < r < b,
(iii) current I out of the page for b < r < c.
Assume each conductor to have a uniform current density. Find B for
(a) r < a,
(b) a < r < b,
(c) b < r < c,
(d) r > c.
Solution:
| Concepts, principles, relations
that apply to the problem: Amperes law |
| Why do they apply? Because of the cylindrical symmetry we can find the magnetic field from Ampere's law alone. |
| How do they apply? (a) r < a: B(r) = m0I(r2/a2)/(2pr)
= m0Ir/(2pa2),
B(r) = B(r) |
| Details of the calculation: None |
Problem 3:
Two concentric spherical shells of radii a (inner) and b (outer) are separated by a material of conductivity s. If they are maintained at a potential difference V, what current flows between them? What is the resistance between the shells when b >> a?
Solution:
| Concepts, principles, relations
that apply to the problem: Gauss' law, symmetry, steady currents |
| Why do they apply? Ñ×j = 0 between the shells. j = sE. Ñ×E = 0 between the shells. |
| How do they apply? Let the inner shell have the higher potential, then the direction of the electric field is radially outward. E = A/r2, V = A(1/a - 1/b) = A(b - a)/(ab), A = Vab/(b - a). j = sVab/[(b - a)r2]. I = 4pr2j = 4psVab/(b - a). R = V/I = (b - a)/(4psVab). When b >> a then R @ 1/(4psVa). |
| Details of the calculation: None= |
Problem 4:
In the circuit shown below, all three voltmeters are ideal and identical. Each resistor has the same given resistance R. Voltage V is also given. Find the reading of each voltmeter.
Solution:
| Concepts, principles, relations
that apply to the problem: Resistors in parallel and in series, model of an ideal voltmeter |
| Why do they apply? We assume the voltmeters are ideal voltmeters and no current passes through the meters. Then we have the points 1 and 2 are at potential V/3 and the point 3 is at potential 2V/3.
We can then model the voltmeters in one of the following
ways. |
| How do they apply? (a) Each voltmeter is represented by a capacitor.
The voltage across the capacitor network is V’ = V/3. The
total capacitance is C’. Each capacitor has capacitance C. 1/C’ = 1/C +
1/(2C). C’ = 2C/3.
|
| Details of the calculation: None |
Problem 5:
A converging magnetic field is often used as a magnetic mirror. Consider a symmetric converging field with ¶Bz/¶z = f(z). Show that the radial component of B in cylindrical coordinates, namely Br, where r = xi + yj is given by Br = -(r/2)f(z).
Solution:
| Concepts, principles, relations that apply to the
problem: Gauss’s law for B |
| Why do they apply? Ñ×B = 0, ¶Bz/¶z = f(z) is given, the field has no f dependence, we can solve for Br. |
| How do they apply? Ñ×B = 0, ¶Bx/¶x + ¶By/¶y = -¶Bz/¶z = -f(z). In cylindrical coordinates: (1/r)¶(rBr)/¶r + (1/r) ¶Bf/¶f = -f(z). For the symmetric converging field Bf = 0, so ¶(rBr)/¶r = - rf(z), rBr = - r2f(z)/2, Br = - rf(z)/2. |
| Details of the calculation: None |
Problem 6:
(a) An aluminum wire has a resistance of 0.10 W. If
you draw this wire through a die, making it thinner and twice as long, what will
be its new resistance?
(b) Four copper wires of equal length are connected in series. The
cross sectional areas are 1 cm2, 2 cm2, 3 cm2,
and 5 cm2. A voltage of 120 V is applied to the arrangement.
What is the voltage across the 2 cm2 wire in units of volt?
Solution:
| Concepts, principles, relations that apply to the
problem: Resistance of a wire |
| Why do they apply? The resistivity of the material does not change. As the shape of the wire changes, its resistance changes. |
| How do they apply? (a) The initial resistance Ri of the aluminum wire with length L and cross-sectional area A is equal to is Ri = rL/A. The initial volume of the wire is AL. After passing the wire through the die, its length has changed to L' and its cross-sectional area is equal A'. Its final volume is therefore equal to A'L'. Since the density of the aluminum does not change, the volume of the wire does not change, and therefore the initial and final dimensions of the wire are related by AL = A'L'. We have A' = L/L'A = A/2. The final resistance Rf of the wire is given by Rf = rL'/A' = 4rL/A = 4Ri = 0.40 W. |
| Details of the calculation: (b) We have 4 segments in series. For each of the segments we have Ri = rl/Ai, where r is the resistivity, l is the length and Ai is the cross-sectional area of the wire. (Here r and l is the same for all wires, but Ai is different for each of them.) For resistors in series we have R = R1 + R2 + R3 + R4 = 104rl(1 + (1/2) + (1/3) + (1/5)) = 104rl 61/30. (area in m2) The voltage is given, V = IR, I = V/R now yields the current flowing through the wires in terms of r and l. I = (120*30 V)/(104 rl 61) The voltage drop across resistor 2 is delta V2 = IR2 = (120´30 V)/(104 rl 61) R2. With R2 = 104 rl(1/2), r and l cancel and V2 = 29.5 V. |
Problem 7:
A particle with mass M and charge q > 0 moves in a uniform magnetic field B and also in the field of another charge Q < 0 located at the origin. At t = 0 the particle is at x = z = 0, y = a, and its velocity is v0i. For what B will the trajectory of the particle be a circle of radius a centered at the origin?
Solution:
| Concepts, principles, relations that apply to the
problem: The Lorentz force |
| Why do they apply? The Lorentz force, F = q(E + v´B) provides the centripetal force required for a circular orbit. |
| How do they apply? Consider the geometry shown in the figure below. Let B = Bk.  For a circular orbit of radius a we need Mv02/a = keq|Q|/a2 + qv0B, B = Mv0/qa - ke|Q|/v0a2. If B is negative, then B points into the page |
| Details of the calculation: None |
