Worked examples · Sheet 2

Fields of loops and circular arcs

Theory used: Field of a circular loop on its axis Field at the center of a circular arc Field of an infinite straight wire Field of a straight segment Direction and sense of the field

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A circular loop of radius $R = 5.0 cm$ carries the current $I = 4.0 A$ . Compute the magnitude of the magnetic field at the center of the loop and at the point on its axis at distance $z = 12 cm$ from the center.

B_{0} = \frac{μ _{0} I}{2 R}, B (z) = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + z ^{2} ) ^{\frac{3}{2}}} = B_{0} \frac{R ^{3}}{( R ^{2} + z ^{2} ) ^{\frac{3}{2}}}

B_{0} R^{2} + z^{2} B (z) = \frac{4 π \cdot 1 0 ^{- 7} \cdot 4.0}{2 \cdot 0.050} T \approx 5.03 \cdot 1 0^{- 5} T = 5. 0^{2} + 1 2^{2} cm = 13 cm = 5.03 \cdot 1 0^{- 5} \cdot \frac{5 ^{3}}{1 3 ^{3}} T = 5.03 \cdot 1 0^{- 5} \cdot \frac{125}{2197} T \approx 2.9 \cdot 1 0^{- 6} T

We apply the theorem on the field of a circular loop on its axis. At the center the formula reduces to $\frac{μ _{0} I}{2 R}$ , and the field is about $5.0 \cdot 1 0^{- 5} T$ . Writing the field on the axis as $B_{0}$ times a dimensionless factor simplifies the calculation: with $R = 5 cm$ and $z = 12 cm$ the distance of the point from the loop is $13 cm$ , because $5$ , $12$ , $13$ form a Pythagorean triple. The factor equals $\frac{125}{2197} \approx 0.057$ , so at $12 cm$ from the center the field is about $6 %$ of the field at the center.

As a check, for $z ≫ R$ the field should approach $\frac{μ _{0} I R ^{2}}{2 z ^{3}} \approx 3.6 \cdot 1 0^{- 6} T$ ; here $z$ is only $2.4$ times $R$ , and the exact value is still noticeably smaller.

Exercise 2

A plane circuit consists of two concentric semicircles, of radii $a = 4.0 cm$ and $b = 8.0 cm$ , connected by two straight segments along the diameter, and carries the current $I = 6.0 A$ , which flows counterclockwise on the outer semicircle. Compute the magnetic field at the common center $C$ .

Circuit made of two semicircles and two radial segments: the current travels the outer arc counterclockwise and the inner arc clockwise

B_{out} = \frac{μ _{0} I π}{4 π b} = \frac{μ _{0} I}{4 b} (⊙), B_{in} = \frac{μ _{0} I}{4 a} (\otimes), B_{rad} = 0

B = B_{in} - B_{out} = \frac{μ _{0} I}{4} (\frac{1}{a} - \frac{1}{b})

B = \frac{4 π \cdot 1 0 ^{- 7} \cdot 6.0}{4} \cdot (\frac{1}{0.040} - \frac{1}{0.080}) T = 1.885 \cdot 1 0^{- 6} \cdot 12.5 T \approx 2.4 \cdot 1 0^{- 5} T

The circuit consists of arcs centered at $C$ and of radial segments, so we apply the theorem on the field at the center of a circular arc to each semicircle, with $φ = π$ . The straight segments lie on the diameter, that is they are aligned with $C$ ; for them $d l$ is parallel to $r$ , and their contribution is zero.

The current travels the closed circuit in a single sense: if it flows counterclockwise on the outer arc, on the left radial segment it goes toward the center and on the inner arc it flows clockwise. By the right-hand rule the outer arc produces at $C$ a field out of the page and the inner arc a field into the page, and the two contributions subtract. The inner arc is closer to the center, therefore its contribution prevails: the field at $C$ is about $2.4 \cdot 1 0^{- 5} T$ and points into the page.

Exercise 3

A very long wire is bent into a hairpin: two parallel semi-infinite straight segments are connected by a semicircle of radius $R = 2.0 cm$ , and the straight segments are tangent to the semicircle at the endpoints of its diameter. The wire carries the current $I = 5.0 A$ . Compute the magnetic field at the center $C$ of the semicircle.

Hairpin wire: the two semi-infinite segments and the semicircle produce at

C

fields out of the page

B_{arc} = \frac{μ _{0} I π}{4 π R} = \frac{μ _{0} I}{4 R}, B_{top} = B_{bottom} = \frac{μ _{0} I}{4 π R}

B = \frac{μ _{0} I}{4 R} + 2 \frac{μ _{0} I}{4 π R} = \frac{μ _{0} I}{4 R} (1 + \frac{2}{π})

B = \frac{4 π \cdot 1 0 ^{- 7} \cdot 5.0}{4 \cdot 0.020} \cdot (1 + 0.637) T = 7.85 \cdot 1 0^{- 5} \cdot 1.637 T \approx 1.3 \cdot 1 0^{- 4} T

The wire is divided into three parts, and by the superposition principle the field at $C$ is the sum of their contributions. The semicircle has center $C$ and angle $π$ , so by the theorem on the field at the center of an arc it gives $\frac{μ _{0} I}{4 R}$ . Each straight segment is a semi-infinite wire ending at the foot of the perpendicular drawn from $C$ , because it is tangent to the semicircle at an endpoint of the diameter; by the theorem on the field of a straight wire its contribution is $\frac{μ _{0} I}{4 π R}$ , half that of an infinite wire.

It remains to compare the senses. On the upper segment the current goes to the left and $C$ is below the wire; on the lower segment the current goes to the right and $C$ is above it; on the semicircle the current circulates counterclockwise. By the right-hand rule all three contributions point out of the page, so the magnitudes add up and the field is about $1.3 \cdot 1 0^{- 4} T$ . Without the straight segments the field would be $61 %$ of this value.

Exercise 4

Two coaxial circular coils, each of $N = 50$ turns of radius $R = 10 cm$ , are placed at distance $R$ from each other and carry the current $I = 2.0 A$ in the same sense (Helmholtz coils). Compute the magnetic field at the point on the axis equidistant from the two coils and compare it with the field at the center of a single coil.

B = 2 \cdot \frac{N μ _{0} I R ^{2}}{2 ( R ^{2} + \frac{R ^{2}}{4} ) ^{\frac{3}{2}}} = \frac{N μ _{0} I}{R} (\frac{4}{5})^{\frac{3}{2}}, B_{1} = \frac{N μ _{0} I}{2 R}

\frac{N μ _{0} I}{R} B B_{1} = \frac{50 \cdot 4 π \cdot 1 0 ^{- 7} \cdot 2.0}{0.10} T \approx 1.257 \cdot 1 0^{- 3} T = 1.257 \cdot 1 0^{- 3} \cdot 0.7155 T \approx 9.0 \cdot 1 0^{- 4} T = \frac{1.257 \cdot 1 0 ^{- 3}}{2} T \approx 6.3 \cdot 1 0^{- 4} T

The equidistant point is on the axis at distance $z = \frac{R}{2}$ from each coil. By the theorem on the field of a circular loop on its axis, multiplied by the number of turns, each coil produces there a field directed along the axis; the currents have the same sense, so the two fields have the same sense and add up. The denominator contains $R^{2} + \frac{R ^{2}}{4} = \frac{5}{4} R^{2}$ , and simplifying we obtain the factor $(\frac{4}{5})^{\frac{3}{2}} \approx 0.716$ .

The field at the midpoint is about $9.0 \cdot 1 0^{- 4} T$ , that is $1.43$ times the field at the center of a single coil. The distance $R$ between the coils is chosen so that the field near the midpoint is nearly uniform, and for this reason Helmholtz coils are used to produce uniform fields in the laboratory.