Problem#1
A straight conductor carrying current i = 5.0 A splits into identical semicircular arcs as shown in Fig. 1. What is the magnetic field at the center C of the resulting circular loop?
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| Fig.1 |
Answer:
The magnitude of the magnetic field at the center of a circular arc, of radius R and central angle φ (in radians), carrying current i, is
B = μ0iφ/4πR
We use the right hand rule to determine the B field direction of the two arcs:
B = Bupper - Blower = 0
Problem#2
In Fig. 2, a current i =10 A is set up in a long hairpin conductor formed by bending a wire into a semicircle of radius R = 5.0 mm. Point bis midway between the straight sections and so distant from the semicircle that each straight section can be approximated as being an infinite wire. What are the (a) magnitude and (b) direction (into or out of the page) of B at a and the (c) magnitude and (d) direction B of at b?
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| Fig.2 |
Answer:
For infinite straight wire: B = μ0i/2πR
For semi─infinite straight wire: B = μ0i/4πR
Center of circular are: B = μ0iφ/4πR
(a) the magnitude of B at a is from the two wire and the circular are
Bnet = 2(μ0i/4πR) + μ0iφ/4πR
Bnet = 2(μ0i/4πR) + μ0iπ/4πR
Bnet = [μ0i/2R][1/2 + 1/π]
= [(4π x 10─7 T.m/A)(10 A)/(2 x 0.005 m)][1/2 + 1/π]
Bnet = 1.0 x 10─7 T
(b) The right-hand rule shows that this field is out the page.
(c) the magnitude of B at b is from the two wire are (we can treat wires as infinite)
Bnet = 2(μ0i/2πR)
Bnet = μ0i/πR
= (4π x 10─7 T.m/A)(10 A)/(2 x 0.005 m)
Bnet = 8.0 x 10─4 T
(b) The right-hand rule shows that this field is out the page.
Problem#3
In Fig. 3, two circular arcshave radius a = 13.5 cm and b = 10.7 cm, subtend angle θ = 74.0°, carry current i = 0.411 A, and share the same center of curvature P. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at P?
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| Fig.3 |
Answer:
We will break the given shape into four pieces. Then, for each piece, we find its contribution to the magnetic field at P.
The net magnetic field Bnet at P it simply the (vector) sum of the (four) contributions. Recall that radial segments (1 and 3) do not contribute any magnetic field.
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| Fig.4 |
Therefore, we only need to focus on 2 and 4. They are circular arcs whose magnetic fields at the center ca be found by
B = μ0iθ/4πR
The directions of the magnetic fields are determined by the curled─straight right─hand rule.
Applying the rule, we found that at P, arc 2 gives B2 which points into the page and arc 4 gives B2 which points out the page.
(a) the net B is given by
Bnet = B4 ─ B2 = μ0iθ/4πb ─ μ0iθ/4πa
Bnet = [μ0iθ/4π][1/b ─ 1/a]
Bnet = [(4π x 10─7 T.m/A)(0.411 A)(74.0° x π/180o)/4π][1/0.107 m ─ 1/0.135 m]
Bnet = 1.03 x 10─7 T
(b) note that B4 > B2, hence the net B will point out of the page
Problem#4
A current is set up in a wire loop consisting of a semicircle of radius 4.00 cm, a smaller concentric semicircle, and two radial straight lengths, all in the same plane. Figure 4a shows the arrangement but is not drawn to scale. The magnitude of the magnetic field produced at the center of curvature is 47.25μT. The smaller semicircle is then flipped over (rotated) until the loop is again entirely in the same plane (Fig. 4b).The magnetic field produced at the (same) center of curvature now has magnitude 15.75 μT, and its direction is reversed from the initial magnetic field. What is the radius of the smaller semicircle?
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| Fig.4 |
Answer:
The radial sections contribute nothing to the field at the center - only the arcs contribute. In configuration (a), the fields at the center due to each semicircle point in the same direction and are thus additive. In configuration (b), the fields oppose eachother so they subtract. Let the large radius and smaller unknown radius be R = 4.00cm and r respectively. Let also the magnitudes of the fields produced by each semicirlce be BR and Br. We have
47.25µT = BR + Br
15.75µT = BR – Br
We know the general form for the B due to a circular arc of radius R subtending angle φ
Bi = μ0iφ/4πR
We can now express BR and Br using φ = π for half circles
47.25µT = [μ0i/4][1/R + 1/r]
15.75µT = [μ0i/4][1/R ─ 1/r]
We don’t know the current i, so let’s get rid of it but taking the ratio of these equations. The left side becomes
47.25/15.75 = [1/R + 1/r]/[1/R ─ 1/r] = 3
1/R + 1/r = 3/R ─ 3/r
2r = R
r = R/2 = 4.00 cm/2 = 2.00 cm
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