Question

Two parallel wires are 5.00 cm apart and carry currents in opposite directions, as shown in Fig. E28.12. 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.

Short answer

The magnetic field at point P due to two 1.50 mm segments is${\mathrm{dB}}_{\mathrm{tot}}=2.64\times {10}^{-7}\mathrm{T}$

Step-by-step solution

Biot savart law

Consider the two parallel wires which are 5.00 cm apart and carry currents in opposition directions. We need to find the magnitude and direction field at point P which at distance 2.50cm from each one of the two-wire due to two segments of wire that are opposite each other and each r = 8.00cm from the point P .The magnetic field due to a segment of length dl and current of I is given by Biot Savart law as:

$\mathrm{dB}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{ldlsin\varphi }}{{\mathrm{r}}^2}$

field at the point P

When the$\varphi$ angle between the position of unit vector and the length of the segment, as shown in the textbook figure this angle is the angle that the dotted line makes with the horizontal axis, we can see from the figure that the$\sin \varphi =2.50cm/8.00cm$ . The magnetic field due to the two segments is into the page. Now, the magnetic field due to these segments at the point P will be:

$d{B}_{tot}=\frac{{\mu }_0}{4\pi }\frac{ldl\sin \varphi }{r^2}+\frac{{\mu }_0}{4\pi }\frac{ldl\sin \varphi }{r^2}$

Now, the current differs in the two wires as:

${\mathrm{dB}}_{\mathrm{tot}}=\frac{{\mathrm{mu}}_0}{4\mathrm{\pi }}\frac{\mathrm{ldlsin\varphi }}{{\mathrm{r}}^2}\left({\mathrm{l}}_1+{\mathrm{l}}_2\right)$

Now putting the values, we get:

$2.64\times {10}^{-7}\mathrm{T}$

Which the magnetic field at the point.