Question

question b

4. (a) State the Bio-Savart law. (1 mark) (b) Use the Bio-Savart law to calculate the magnetic field at a point, P, a distance r from a finite wire segment (see figure), of length 21, carrying a steady current I. Hence determine the magnetic field at the centre of a square loop, side-length 2L, carrying a steady current I. Р. (5 marks) -L 0 +L dx 1 You may need that 2 a2 3 (Vx2 + a2) a2 1+ L2 (c) Modify the derivation in part (b) to find the magnetic field a distance r from an infinitely-long wire carrying a steady current I. (1 mark) +00 dx You may need that s (Vx2 + a2) 2 a2 3

Answer 1

b) see the diagram:

dx > Khaota -L D AL dan figol

Refer to the fig. 1:

Considering an element at a distance x from center of length dx.

$sin\theta = \frac{L}{\sqrt{r^{2}+L^{2}}}$

$x=r\tan\alpha$

------(i)

2 dx = r sec * da

also,

coaa х2 + r2

Apply Biot-Savart's Law:

magnetic field due to the element is

$dB = \frac{\mu_{o}i*dx*sin(90^{\circ}-\alpha)}{4\pi (\sqrt{r^{2}+x^{2}})^{2}}$

The net magnetic field is

$B = \int_{-\theta}^{\theta}\frac{\mu_{o}i*dx*cos(\alpha)}{4\pi (\sqrt{r^{2}+x^{2}})^{2}}$

$= \int_{-\theta}^{\theta}\frac{\mu_{o}i*cos(\alpha)*r*\sec^{2}\alpha*d\alpha}{4\pi (\sqrt{r^{2}+x^{2}})^{2}}$

$= \int_{-\theta}^{\theta}\frac{\mu_{o}i*cos(\alpha)*r^{2}*\sec^{2}\alpha*d\alpha}{4\pi r(\sqrt{r^{2}+x^{2}})^{2}}$

$= \int_{-\theta}^{\theta}\frac{\mu_{o}i*cos(\alpha)*cos^{2}\alpha*\sec^{2}\alpha*d\alpha}{4\pi r}$

$= \frac{\mu_{o}i}{4\pi r}\int_{-\theta}^{\theta}cos(\alpha)*d\alpha$

$= \frac{\mu_{o}i}{4\pi r}*2sin\theta$

$= \frac{\mu_{o}i}{2\pi r}*\frac{L}{\sqrt{r^{2}+L^{2}}}$ [answer]

Refer to the fig. 2:

CA Q345° k 17-1 2.

The length of each side is 2L.

the distance of the center from the side of the square is r = L.

The magnetic field due to one side is

$B=\frac{\mu_{o}i}{2\pi r}*\frac{L}{\sqrt{r^{2}+L^{2}}}$ [using the result above]

As the direction of the magnetic field due to all the sides at the center is the same and has the same magnitude.

Therefore, the net magnetic field at the center is $B_{net}=4B=4*\frac{\mu_{o}i}{2\pi *L}*\frac{L}{\sqrt{L^{2}+L^{2}}}$

  $=1*\frac{\mu_{o}i}{\pi}*\frac{1}{L\sqrt{2}}$

  $=\frac{\mu_{o}i}{\sqrt{2}\pi L}$ [answer]