12. Describe the principle of the direct current motor
13. How far from a straight current−carrying wire carrying a
current of 1 A is the strength of the magnetic field 1.5 × 10−6 T?
The permeability of free space is 4π × 10−7 T m/A.
14. Show how to derive the expression for the magnetic field
between two wires.
15. Find the contribution to the magnetic field at a point that is a
perpendicular distance 15 cm from a current element of length
20 cm, where the current through the element is 5 A. The
permeability of free space is 4π × 10−7 T m/A.
16. State Ampere’s law.
17. How far from the centre of a conductor of radius 10 mm with
a current of 1.5 A flowing through it is the magnetic field
strength 5 × 10−6 T? The permeability of free space is 4π × 10−7
T m/A.
18. Consider the solenoid shown in Figure 6.28
B
I I
Detail of bottom of Ampere's law path
rectangle inside solenoid
Figure 6.28
Derive the expression for the magnetic field of this solenoid.
19. What current is needed through a solenoid of 350 turns if the
magnetic field inside it is to be 9 × 10−4? The permeability of
free space is 4π × 10−7 T m/A.
20. Find the magnetic field for a toroid of radius 3 cm with 200
turns per unit length with a current of 2 A flowing through it.
The permeability of free space is 4π × 10−7 T m/A.
21. What is the range of the strength of the magnetic field at the
earth’s surface.
22. Describe how a tangent galvanometer works.�
12. The principle of the direct current motor is that a current-carrying conductor placed in a magnetic field experiences a force and begins to rotate.
13. Using the formula for the magnetic field around a straight current-carrying wire, B = (μ₀I)/(2πr), where μ₀ is the permeability of free space, I is the current, and r is the distance from the wire, we can plug in the given values to get:
B = (4π × 10−7 T m/A)(1 A)/(2π(1.5 × 10−6 m)) = 0.106 T
So the strength of the magnetic field is 0.106 T, or 106 µT.
14. The expression for the magnetic field between two wires can be derived using Ampere's Law. We can imagine a closed loop around one wire and apply Ampere's Law, which states that the line integral of the magnetic field around a closed loop is equal to the current passing through the loop. Solving for the magnetic field gives:
B = (μ₀I)/(2πr)
where I is the current, r is the distance between the wires, and μ₀ is the permeability of free space.
15. Using the formula for the magnetic field around a current element, B = (μ₀I)/(4πr), where I is the current through the element, r is the distance from the element, and μ₀ is the permeability of free space, we can plug in the given values to get:
B = (4π × 10−7 T m/A)(5 A)/(4π(0.15 m)) = 0.0267 T
So the contribution to the magnetic field at the given point is 0.0267 T, or 26.7 mT.
16. Ampere's Law states that the line integral of the magnetic field around a closed loop is equal to the current passing through the loop multiplied by the permeability of free space:
∫ B · dl = μ₀I
where B is the magnetic field, dl is a small element of length along the loop, I is the current passing through the loop, and μ₀ is the permeability of free space.
17. Using the formula for the magnetic field around a conductor, B = (μ₀I)/(2πr), where I is the current through the conductor, r is the distance from the centre of the conductor, and μ₀ is the permeability of free space, we can plug in the given values to get:
5 × 10−6 T = (4π × 10−7 T m/A)(1.5 A)/(2πr)
Solving for r gives:
r = (4π × 10−7 T m/A)(1.5 A)/(2π(5 × 10−6 T)) = 0.029 m
So the distance from the centre of the conductor is 2.9 cm.
18. The magnetic field of a solenoid can be derived using Ampere's Law. If we imagine a rectangular loop of wire that encloses n turns of wire inside the solenoid, the line integral of the magnetic field around the loop can be simplified to:
B · 2l = μ₀nI
where B is the magnetic field, l is the length of the rectangle (which is also the length of the solenoid), n is the number of turns per unit length, I is the current through the wire, and μ₀ is the permeability of free space. Solving for B gives:
B = (μ₀nI)/(2l)
19. Using the formula for the magnetic field of a solenoid, B = (μ₀nI), where n is the number of turns per unit length and I is the current through the solenoid, we can plug in the given values and solve for I:
9 × 10−4 T = (4π × 10−7 T m/A)(350 turns/m)I
Solving for I gives:
I = (9 × 10−4 T)/(4π × 10−7 T m/A)(350 turns/m) = 2.04 A
So the current needed through the solenoid is 2.04 A.
20. The magnetic field of a toroid can be derived using the formula B = (μ₀nI), where n is the number of turns per unit length and I is the current through the toroid, and by using the formula for the magnetic field inside a solenoid, B = (μ₀nI) for a toroidal solenoid. Plugging in the given values gives:
B = (4π × 10−7 T m/A)(200 turns/m)(2 A)/(2π(0.03 m)) = 0.84 T
So the magnetic field is 0.84 T.
21. The strength of the magnetic field at the earth's surface can range from about 25 µT to 65 µT, depending on location.
22. A tangent galvanometer works by using the magnetic field generated by a current in a coil of wire to deflect a compass needle. The current passing through the coil creates a magnetic field, which interacts with the Earth's magnetic field to deflect the compass needle. By adjusting the current, the angle of deflection can be measured and used to calculate the strength of the magnetic field.