Problem#1
A short current element dl = (0.500 mm)j carries a current of 8.20 A in the same direction as dl. Point P is located at r = (─0.730 m)i+ (0.390 m)k. Use unit vectors to express the magnetic field at P produced by this current element.Answer:
The law of Biot and Savart gives
dB = (μ0/4π)(Idl x r/r3)
known:
element dl = (0.500 mm)j
carries a current, I = 8.20 A
Point P is located at r = (─0.730 m)i+ (0.390 m)k.
Then,
r = [(─0.730 m)2 + (0.390 m)2]1/2 = 0.828 m
therefore:
dB = (4π x 10─7 Tm/A)(8.20 A)(0.500 x 10─3 m)j x [(─0.730 m)i+ (0.390 m)k]/[4π x (0.828 m)3)
dB = (7.22 x 10─10j) x [(─0.730 m)i+ (0.390 m)k]
= (5.27 x 10─10)k + (2.82 x 10─10)i
dB = (2.82i + 5.27k) x 10─10T
(Note: j x i = ─k, and j x k = i)
Problem#2
A straight wire carries a 10.0-A current (Fig. 1). ABCD is a rectangle with point D in the middle of a 1.10-mm segment of the wire and point C in the wire. Find the magnitude and direction of the magnetic field due to this segment at (a) point A; (b) point B; (c) point C.
 |
| Fig.1 |
The law of Biot and Savart gives
dB = (μ0/4π)(Idl sin φ/r2)
(a) the magnitude and direction of the magnetic field due to this segment at point A is (with dl = 1.10 mm = 1.10 x 10─3 m, φ = 900, r = 5.00 cm = 5.00 x 10─2 m)
dB = (4π x 10─7 Tm/A)(10.0 A)(1.10 x 10─3 m) sin 900/[4π x (5.00 x 10─2 m)2]
dB = 4.40 x 10─7 T, out of the paper.
(b) the magnitude and direction of the magnetic field due to this segment at point B.
dB = (μ0/4π)(Idl sin φ/rB2)
with dl = 1.10 mm = 1.10 x 10─3 m
r = [(5 cm)2 + (14 cm)2]1/2 = 14.87 cm = 14.87 x 10─2 m
angle, φ = arc tan (5/14) = 19.650
therefore:
dB = (4π x 10─7 Tm/A)(10.0 A)(1.10 x 10─3 m) sin 19.650/[4π x (14.87 x 10─2 m)2]
dB = 1.67 x 10─8 T, out of the paper.
(a) the magnitude and direction of the magnetic field due to this segment at point A is dB = 0, since φ = 0.
Problem#3
A long, straight wire, carrying a current of 200 A, runs through a cubical wooden box, entering and leaving through holes in the centers of opposite faces (Fig. 2). The length of each side of the box is 20.0 cm. Consider an element dl of the wire 0.100 cm long at the center of the box. Compute the magnitude dB of the magnetic field produced by this element at the points a, b, c, d, and e in Fig. E28.12. Points a, c, and d are at the centers of the faces of the cube; point b is at the midpoint of one edge; and point e is at a corner. Copy the figure and show the directions and relative magnitudes of the field vectors. (Note: Assume that the length dl is small in comparison to the distances from the current element to the points where the magnetic field is to be calculated.)
 |
| Fig.2 |
The magnitude of the field due to the current element is
dB = (μ0/4π)(Idl sin φ/r2)
Where φ is angle between r and the current direction.
 |
| Fig.3 |
The magnetic field at the given points is:
For point a: (with ra = 0.100 m, φ = 900)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m) sin 900/[4π x (0.100 m)2]
dB = 2.00 x 10─7 T
For point b: (with rb = 0.100√2 m , φ = 450)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m) sin 450/[4π x (0.100√2 m)2]
dB = 7.07 x 10─8 T
For point c: (with rc = 0.100 m , φ = 900)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m) sin 900/[4π x (0.100 m)2]
dB = 2.00 x 10─7 T
For point d: (with rc = 0.100 m , φ = 00)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m) sin 00/[4π x (0.100 m)2]
dB = 0
For point e: (with re = 0.100√3 m , sin φ = √2/√3)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m)(√2/√3)/[4π x (0.100√3 m)2]
dB = 5.44 x 10─8 T
Problem#4
A long, straight wire lies along the z-axis and carries a 4.00-A current in the +z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire centered at the origin: (a) x = 2.00 m, y = 0, z = 0; (b) x = 0, y = 2.00 m, z = 0; (c) x = 2.00 m, y = 2.00 m, z = 0; (d) x = 0, y = 0, z = 2.00 m.
Answer:
The law of Biot and Savart gives
dB = (μ0/4π)(Idl x r/r3)
known:
element dl = (0.500 mm)k = (0.500 x 10─3 m)k (because I is in the +z─direction)
carries a current, I = 4.00 A
(a) Field oint is at x = 2.00 m, y = 0, z = 0 so vector r from the source point (at the origin) to the field point is r = (2.00 m)i.
therefore:
dB = (4π x 10─7 Tm/A)(4.00 A) (0.500 x 10─3 m)k x (2.00 m)i/[4π x (2.00 m)3]
dB = (5.00 x 10─11 T)j
(b) Field oint is at x = 0, y = 2.00 m, z = 0 so vector r from the source point (at the origin) to the field point is r = (2.00 m)j.
therefore:
dB = (4π x 10─7 Tm/A)(4.00 A) (0.500 x 10─3 m)k x (2.00 m)j/[4π x (2.00 m)3]
dB = ─(5.00 x 10─11 T)i
(c) Field oint is at x = 2.00 m, y = 2.00 m, z = 0 so vector r from the source point (at the origin) to the field point is r = (2.00 m)(i + j) m, r = √2 x 2.00 m
therefore:
dB = (4π x 10─7 Tm/A)(4.00 A) (0.500 x 10─3 m)k x (2.00 m)(i + j)m/[4π x (√2 x 2.00 m)3]
dB = ─(1.77 x 10─11 T)(i ─ j)
(d) Field oint is at x = 0, y = 0, z = 2.00 m so vector r from the source point (at the origin) to the field point is r = (2.00 m)k and r = 2.00 m
therefore:
dB = (4π x 10─7 Tm/A)(4.00 A) (0.500 x 10─3 m)k x (2.00 m)k/[4π x (2.00 m)3]
dB = ─(5.00 x 10─11 T)(k x k) = 0
Problem#5
Two parallel wires are 5.00 cm apart and carry currents in opposite directions, as shown in Fig. 5. Find the magnitude and direction of the magnetic field at point P due to two 1.50-mm segments of wire that are opposite each other and each 8.00 cm from P.
Answer:
The magnitude of the field due to the current element is
dB = (μ0/4π)(Idl sin φ/r2)
Where φ is angle between r and the current direction.
Applying the law Biot and Savart for the 12.0 A current gives
dB = (4π x 10─7 Tm/A)(12.0 A)(1.50 x 10─3 m)(2.50 cm/8.00 cm)/[4π x (0.0800 m)2]
dB = 8.79 x 10─8 T, into the page
Applying the law Biot and Savart for the 24.0 A current gives
dB = (4π x 10─7 Tm/A)(24.0 A)(1.50 x 10─3 m)(2.50 cm/8.00 cm)/[4π x (0.0800 m)2]
dB = 1.76 x 10─7 T, into the page
so, the total field is
dB = 2.64 x 10─7 T, into the page
Problem#6
A wire carrying a 28.0-A current bends through a right angle. Consider two 2.00-mm segments of wire, each 3.00 cm from the bend (Fig. 6). Find the magnitude and direction of the magnetic field these two segments produce at point P, which is midway between them.
Answer:
The magnitude of the field due to the current element is
dB = (μ0/4π)(Idl sin φ/r2)
Where φ is angle between r and the current direction.
Applying the Biot and Savart law, where
r = ½ [(3.00 cm)2 + (3.00 cm)2]1/2 = 2.121 cm, we have
dB = 2(4π x 10─7 Tm/A)(28.0 A)(2.00 x 10─3 m) sin 450/[4π x (0.02121 m)2]
dB = 1.76 x 10─5 T, into the page
Problem#7
A square wire loop 10.0 cm on each side carries a clockwise current of 15.0 A. Find the magnitude and direction of the magnetic field at its center due to the four 1.20-mm wire segments at the midpoint of each side.
Answer:
The magnitude of the field due to the current element is
dB = (μ0/4π)(Idl sin φ/r2)
Where φ is angle between r and the current direction.
For point a: (with ra = 0.100 m, φ = 900)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m) sin 900/[4π x (0.100 m)2]
dB = 2.00 x 10─7 T
For point b: (with rb = 0.100√2 m , φ = 450)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m) sin 450/[4π x (0.100√2 m)2]
dB = 7.07 x 10─8 T
For point c: (with rc = 0.100 m , φ = 900)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m) sin 900/[4π x (0.100 m)2]
dB = 2.00 x 10─7 T
For point d: (with rc = 0.100 m , φ = 00)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m) sin 00/[4π x (0.100 m)2]
dB = 0
For point e: (with re = 0.100√3 m , sin φ = √2/√3)
dB = (4π x 10─7 Tm/A)(200 A)(1.00 x 10─4 m)(√2/√3)/[4π x (0.100√3 m)2]
dB = 5.44 x 10─8 T
 |
| Fig.4 |
A long, straight wire lies along the z-axis and carries a 4.00-A current in the +z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire centered at the origin: (a) x = 2.00 m, y = 0, z = 0; (b) x = 0, y = 2.00 m, z = 0; (c) x = 2.00 m, y = 2.00 m, z = 0; (d) x = 0, y = 0, z = 2.00 m.
Answer:
The law of Biot and Savart gives
dB = (μ0/4π)(Idl x r/r3)
known:
element dl = (0.500 mm)k = (0.500 x 10─3 m)k (because I is in the +z─direction)
carries a current, I = 4.00 A
(a) Field oint is at x = 2.00 m, y = 0, z = 0 so vector r from the source point (at the origin) to the field point is r = (2.00 m)i.
therefore:
dB = (4π x 10─7 Tm/A)(4.00 A) (0.500 x 10─3 m)k x (2.00 m)i/[4π x (2.00 m)3]
dB = (5.00 x 10─11 T)j
(b) Field oint is at x = 0, y = 2.00 m, z = 0 so vector r from the source point (at the origin) to the field point is r = (2.00 m)j.
therefore:
dB = (4π x 10─7 Tm/A)(4.00 A) (0.500 x 10─3 m)k x (2.00 m)j/[4π x (2.00 m)3]
dB = ─(5.00 x 10─11 T)i
(c) Field oint is at x = 2.00 m, y = 2.00 m, z = 0 so vector r from the source point (at the origin) to the field point is r = (2.00 m)(i + j) m, r = √2 x 2.00 m
therefore:
dB = (4π x 10─7 Tm/A)(4.00 A) (0.500 x 10─3 m)k x (2.00 m)(i + j)m/[4π x (√2 x 2.00 m)3]
dB = ─(1.77 x 10─11 T)(i ─ j)
(d) Field oint is at x = 0, y = 0, z = 2.00 m so vector r from the source point (at the origin) to the field point is r = (2.00 m)k and r = 2.00 m
therefore:
dB = (4π x 10─7 Tm/A)(4.00 A) (0.500 x 10─3 m)k x (2.00 m)k/[4π x (2.00 m)3]
dB = ─(5.00 x 10─11 T)(k x k) = 0
Problem#5
Two parallel wires are 5.00 cm apart and carry currents in opposite directions, as shown in Fig. 5. Find the magnitude and direction of the magnetic field at point P due to two 1.50-mm segments of wire that are opposite each other and each 8.00 cm from P.
 |
| Fig.5 |
The magnitude of the field due to the current element is
dB = (μ0/4π)(Idl sin φ/r2)
Where φ is angle between r and the current direction.
Applying the law Biot and Savart for the 12.0 A current gives
dB = (4π x 10─7 Tm/A)(12.0 A)(1.50 x 10─3 m)(2.50 cm/8.00 cm)/[4π x (0.0800 m)2]
dB = 8.79 x 10─8 T, into the page
Applying the law Biot and Savart for the 24.0 A current gives
dB = (4π x 10─7 Tm/A)(24.0 A)(1.50 x 10─3 m)(2.50 cm/8.00 cm)/[4π x (0.0800 m)2]
dB = 1.76 x 10─7 T, into the page
so, the total field is
dB = 2.64 x 10─7 T, into the page
Problem#6
A wire carrying a 28.0-A current bends through a right angle. Consider two 2.00-mm segments of wire, each 3.00 cm from the bend (Fig. 6). Find the magnitude and direction of the magnetic field these two segments produce at point P, which is midway between them.
 |
| Fig.6 |
The magnitude of the field due to the current element is
dB = (μ0/4π)(Idl sin φ/r2)
Where φ is angle between r and the current direction.
Applying the Biot and Savart law, where
r = ½ [(3.00 cm)2 + (3.00 cm)2]1/2 = 2.121 cm, we have
dB = 2(4π x 10─7 Tm/A)(28.0 A)(2.00 x 10─3 m) sin 450/[4π x (0.02121 m)2]
dB = 1.76 x 10─5 T, into the page
Problem#7
A square wire loop 10.0 cm on each side carries a clockwise current of 15.0 A. Find the magnitude and direction of the magnetic field at its center due to the four 1.20-mm wire segments at the midpoint of each side.
Answer:
The magnitude of the field due to the current element is
dB = (μ0/4π)(Idl sin φ/r2)
Where φ is angle between r and the current direction.
 |
| Fig.7 |
All four fields are of equal magnitude and into the page, so their magnitudes add.
dBnet = 4(4π x 10─7 Tm/A)(15.0 A)(1.20 x 10─3 m) sin 900/[4π x (0.0500 m)2]
dBnet = 2.88 x 10─6 T, into the page
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