Task Description

Given a straight cable, which is placed on the $z$ -Axis. Assuming a constant current $I$ is flowing from $A$ to $B$ , calculate the magnetic field in a point $P$ . Assume point $A$ lies at distance $a$ from the origin on the $z$ -Axis. Assume point $B$ lies at distance $b$ from the origin on the z-Axis. Assume point $P$ is at distance $p$ from the origin on the $x$ -Axis.

Hint: Biot-Savart law $d\vec B=\frac{\mu_0 \cdot I}{4\pi} \cdot \frac{d\vec l \times \vec u_r}{r^2}$ , where $d\vec B$ is the magnetic field created by an infinitely small cable of length and orientation $d\vec l$ in a point which is at distance $r$ to the infinitely small cable and the unit vector of the direction of the point relative to the cable is $\vec u_r$ . $\mu_0$ is the permeability of free space and $I$ is the current flowing through the cable.

Solution

We can write the points as $A=\begin{pmatrix} 0 \\ 0 \\ a \\ \end{pmatrix}$ , $B=\begin{pmatrix} 0 \\ 0 \\ b \\ \end{pmatrix}$ and $P=\begin{pmatrix} p \\ 0 \\ 0 \\ \end{pmatrix}.$ In order to calculate the magnetic field, we use the Biot-Savart law which states that a infinite small cable segment with a current $I$ creates an infinitely small magnetic field in the form of $d\vec B=\frac{\mu_0 \cdot I}{4\pi} \cdot \frac{d\vec l \times \vec u_r}{r^2}$ .

The unit vector $\vec u_r$ points from a point along the cable (any point, the exact point is given by the angle $\alpha$ ) to the point $P$ , so the $\vec u_r$ will also depend on alpha:

We can write the formula of the magnetic field component wise, because it is a vector, then we can perform the crossproduct to get the formula:$$d\vec B = \begin{pmatrix} dB_x \\ dB_y \\ dB_z \\ \end{pmatrix}=\frac{\mu_0 \cdot I}{4\pi r^2} \cdot \begin{pmatrix} 0 \\ 0 \\ dl \\ \end{pmatrix} \times \begin{pmatrix} \cos\alpha \\ 0 \\ -\sin\alpha \\ \end{pmatrix}=\frac{\mu_0 \cdot I}{4\pi r^2} \begin{pmatrix} 0 \\ \cos\alpha\cdot dl \\ 0 \\ \end{pmatrix}$$ It turns out, that the magnetic field will only have a component in the $y$ direction, the components in $z$ and $x$ direction are always zero. Thus we only have to perform the calculation for the $y$ direction, which is given by $$dB_y=\frac{\mu_0\cdot I}{4\pi r^2}\cdot \cos\alpha \cdot dl$$ We integrate $dB_y$ to get the total magnetic field in the $y$ axis: $$\int dB_y=\int_a^b \frac{\mu_0\cdot I}{4\pi r^2}\cdot \cos\alpha \cdot dl$$ This equation can be rewritten by pulling out all the constant factors (everything that does not depend on the relative position on the cable): $$B_y= \frac{\mu_0\cdot I}{4\pi }\int_a^b \frac{\cos\alpha}{r^2} \,dl.$$ Since $\alpha$ and $r$ depend on $l$ , we need to express them in terms of $l$ . By applying the pythagorean theorem, we know that $r=\sqrt{l^2+d^2}$ . Furthermore we know that $\sin\alpha=\frac{d}{r}=\frac{d}{\sqrt{l^2+d^2}}$ . So we can write the integral as $$B_y= \frac{\mu_0\cdot I}{4\pi }\int_a^b \frac{d}{\sqrt{l^2+d^2}^3} \,dl.$$ As this integral is not easily solvable, we try a different approach, we try not to express $\alpha$ in terms of $l$ , but the other way round. Using simple trigonometry, we can write $l=d\cdot\tan\alpha$ . We take the derivative and get $\frac{dl}{d\alpha}=\frac{d}{\cos^2\alpha}$ . This gives us $$B_y= \frac{\mu_0\cdot I}{4\pi }\int_a^b \frac{\cos\alpha}{r^2} \,dl= \frac{\mu_0\cdot I}{4\pi }\int^{\varphi_b}_{\varphi_a} \frac{\cos\alpha}{r^2}\cdot\frac{d}{\cos^2\alpha} \,d\alpha.$$ Where $\varphi_a$ and $\varphi_b$ are the limits of integration expressed in therms of the angle $\alpha$ . Next we can write $r$ in terms of $\alpha$ as $r=\frac{d}{\cos\alpha}$ . $$B_y= \frac{\mu_0\cdot I}{4\pi }\int^{\varphi_b}_{\varphi_a} \cos\alpha\cdot\frac{\cos^2\alpha}{d^2}\cdot\frac{d}{\cos^2\alpha} \,d\alpha.$$ The $\cos\alpha$ and $d$ terms cancle, which gives us $$B_y= \frac{\mu_0\cdot I}{4\pi\cdot d}\int^{\varphi_b}_{\varphi_a} \cos\alpha \,\,d\alpha.$$ Integrating $\cos\alpha$ gives $\sin\alpha$ , so we get $$B_y= \frac{\mu_0\cdot I}{4\pi\cdot d} \left[ \sin\alpha \right]^{\varphi_b}_{\varphi_a}= \frac{\mu_0\cdot I}{4\pi\cdot d} (\sin\varphi_b-\sin\varphi_a ).$$

Now can calculate the $\sin\varphi_a=\frac{a}{\sqrt{a^2+p^2}}$ and $\sin\varphi_b=\frac{b}{\sqrt{b^2+p^2}}$ . This gives $$B_y=\frac{\mu_0\cdot I}{4\pi\cdot d} (\frac{b}{\sqrt{b^2+d^2}}-\frac{a}{\sqrt{a^2+d^2}} ).$$ So the total magnetic field is given by $$B=\frac{\mu_0\cdot I}{4\pi\cdot d} (\frac{b}{\sqrt{b^2+d^2}}-\frac{a}{\sqrt{a^2+d^2}} ) \begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix} .$$