Worked examples · Sheet 1

Fields of infinite straight wires

Theory used: Field of an infinite straight wire Direction and sense of the field Biot-Savart law

Exercise 1

A very long straight wire carries the current $I = 15 A$ . Compute the magnitude of the magnetic field at distance $d = 3.0 cm$ from the wire and the distance $d^{'}$ at which the field of the wire has magnitude $B_{T} = 5.0 \cdot 1 0^{- 5} T$ , a typical value of the Earth’s magnetic field.

B = \frac{μ _{0} I}{2 π d}, d^{'} = \frac{μ _{0} I}{2 π B _{T}}

B d^{'} = \frac{4 π \cdot 1 0 ^{- 7} \cdot 15}{2 π \cdot 0.030} T = \frac{2 \cdot 1 0 ^{- 7} \cdot 15}{0.030} T = 1.0 \cdot 1 0^{- 4} T = \frac{2 \cdot 1 0 ^{- 7} \cdot 15}{5.0 \cdot 1 0 ^{- 5}} m = 0.060 m = 6.0 cm

The wire is very long compared with the distances considered, so the theorem on the field of an infinite straight wire applies. The factor $\frac{μ _{0}}{2 π}$ equals $2 \cdot 1 0^{- 7} T \cdot m/A$ , and it is convenient to use it directly in the calculations. The field is inversely proportional to the distance, therefore the value $B_{T}$ , half of the one found at $3.0 cm$ , is reached at twice the distance, $d^{'} = 6.0 cm$ . Beyond a few tens of centimeters the field of the wire is therefore much smaller than the Earth’s field.

Exercise 2

Two infinite parallel straight wires, perpendicular to the page, are $D = 10 cm$ apart and carry the currents $I_{1} = 20 A$ and $I_{2} = 10 A$ , both out of the page. Compute the magnetic field at the midpoint $M$ of the segment joining the wires and find the point in the plane of the page where the field is zero.

B_{y} (x) = \frac{μ _{0} I _{1}}{2 π x} - \frac{μ _{0} I _{2}}{2 π ( D - x )}, 0 < x < D

B_{y} (\frac{D}{2}) = \frac{μ _{0}}{2 π} \frac{2 ( I _{1} - I _{2} )}{D} = 2 \cdot 1 0^{- 7} \cdot \frac{2 \cdot ( 20 - 10 )}{0.10} T = 4.0 \cdot 1 0^{- 5} T

\frac{I _{1}}{x} = \frac{I _{2}}{D - x} ⟹ x = \frac{I _{1} D}{I _{1} + I _{2}} = \frac{20 \cdot 0.10}{30} m \approx 0.067 m = 6.7 cm

We place wire $1$ at the origin and wire $2$ at the point $x = D$ of the $x$ axis, with the $y$ axis in the plane of the page. The currents come out of the page, so by the right-hand rule the field lines of each wire circulate counterclockwise. At a point between the two wires the field of wire $1$ , which is on the left, points toward $+ y$ , while the field of wire $2$ , which is on the right, points toward $- y$ ; by the superposition principle the component $B_{y}$ is the difference of the two magnitudes.

At the midpoint the distances are equal, therefore the field of the larger current prevails and the total field equals $4.0 \cdot 1 0^{- 5} T$ , directed toward $+ y$ . The field vanishes where the two magnitudes are equal, that is $6.7 cm$ from wire $1$ and $3.3 cm$ from wire $2$ : the point is closer to the wire with the smaller current. Outside the segment the two fields have the same sense, because the point is on the same side of both wires, and at no point off the line of the wires are the two fields parallel; hence the point found is the only point of zero field.

Exercise 3

Two infinite straight wires, perpendicular to the page, pass through the points $A = (- 4.0; 0) cm$ and $C = (4.0; 0) cm$ and carry the same current $I = 10 A$ , out of the page in wire $1$ , which passes through $A$ , and into the page in wire $2$ , which passes through $C$ . Compute the magnetic field vector at the point $P = (0; 3.0) cm$ .

Fields of the two wires at

P

: each is perpendicular to the line joining

P

to its own wire, and the horizontal components cancel

r_{1} = r_{2} = 4. 0^{2} + 3. 0^{2} cm = 5.0 cm, B_{1} = B_{2} = \frac{μ _{0} I}{2 π r _{1}} = \frac{2 \cdot 1 0 ^{- 7} \cdot 10}{0.050} T = 4.0 \cdot 1 0^{- 5} T

B_{1} = B_{1} \frac{k ^ \times r _{1}}{r _{1}} = B_{1} \frac{- 3  ^ + 4  ^}{5}, B_{2} = B_{2} \frac{- k ^ \times r _{2}}{r _{2}} = B_{2} \frac{3  ^ + 4  ^}{5}

B = B_{1} + B_{2} = 4.0 \cdot 1 0^{- 5} \cdot \frac{8}{5} \overset{}{^} T = 6.4 \cdot 1 0^{- 5} \overset{}{^} T

The point $P$ is $5.0 cm$ from both wires, and the currents are equal, so the two fields have the same magnitude $4.0 \cdot 1 0^{- 5} T$ . The directions are obtained from the right-hand rule: for a wire with the current out of the page the field is $\hat{k} \times r$ normalized, where $r$ is the vector going from the wire to the point, in the plane of the page; for the current into the page the sense is reversed. With $r_{1} = (4 \overset{}{^} + 3 \overset{}{^}) cm$ and $r_{2} = (- 4 \overset{}{^} + 3 \overset{}{^}) cm$ , and with $\hat{k} \times \overset{}{^} = \overset{}{^}$ , $\hat{k} \times \overset{}{^} = - \overset{}{^}$ , we obtain the unit vectors written above.

The components along $x$ are opposite and cancel, while those along $y$ add up. The field at $P$ is therefore vertical, pointing toward $+ y$ , and equals $6.4 \cdot 1 0^{- 5} T$ . The result is consistent with the symmetry: at the midpoint $O$ of $A C$ the two fields both point toward $+ y$ , and along the perpendicular bisector of the segment the field remains parallel to that direction.