Question

[img]https://upload.cc/i1/2020/04/13/smpaLD.jpg[/img]




question in photo , R=4

Answer 1

Problem: $I_1=2A; I_2=9A; P_1=(40,60)cm; P_2=(40,-30)cm;\vec H(25,0)=?$

Solution: For an infinitely long conductor the magnetic field at any point $r$ is determined by the formula $(1) \vec H=\frac{1}{4\pi}\frac{2 I \cdot \hat l ×\hat r}{r}$ where $\hat l$ -unit vector along direction of current in conductor, $\hat r$ -unit vector from conductor to point of field to determine. In the figure 1 of https://upload.cc/i1/2020/04/13/smpaLD.jpg, the direction of current in the conductors is shown by a cross, so the current is directed down from the drawing plane in the direction against the Z axis, so $\hat l=-\hat k$. A magnetic field of system of currents is a vector sum of magnetic fields of every currents. Thus we find the magnetic field of first conductor. Find $\vec r_1=25 \hat i -40\hat i - 60 \hat j=-15 \hat i-60\hat j$

(2) $r_1=\sqrt{15^2+60^2} cm=61.8 cm=0.618 m; \hat r_1=\frac {\vec r_1}{r_1}=-\frac{15 \hat i +60 \hat j}{61.8}=-0.24 \hat i-0.97 \hat j$

Vector product can be calculated using the rule $\hat k ×\hat i=\hat j$ and $\hat k ×\hat j=-\hat i$ . Substitute (2) to (1) we determine the magnetic field of first conductor

(3) $\vec H_1=\frac{1}{4\pi}\frac{2I_1 (-\hat k) ×(-0.24 \hat i-0.97 \hat j)}{r_1}=\frac{1}{4\pi}\frac{4A }{0.618 m}\cdot (-0.97 \hat i+0.24\hat j)=\\=0.515\cdot (-0.97 \hat i+0.24\hat j)Am^{-1}=(-0.500\hat i +0,124 \hat j )Am^{-1}$

For second conductor we have

$\vec r_2=25\hat i -40 \hat i + 30\hat j=-15 \hat i +30 \hat j; r_2=\sqrt{15^2+30^2}=33.5 cm=0.335m$

$\hat r_2=\frac{\vec r_2}{r_2}=\frac{-15 \hat i +30 \hat j}{33.5}=-0.447 \hat i+0.894 \hat j$

(4) $\vec H_2=\frac{1}{4\pi}\frac{2I_2 (-\hat k) ×(-0.447 \hat i+0.894 \hat j)}{r_2}=\frac{1}{4\pi}\frac{18A }{0.335 m}\cdot (0.894 \hat i+0.447\hat j)=\\=4.28\cdot (0.894 \hat i+0.447\hat j)Am^{-1}=(3.82\hat i +1.91 \hat j)Am^{-1}$

The sketch explains the resulting picture of the fields

Now we can get overall magnetic field

$\vec H=\vec H_1+\vec H_2=(-0.500\hat i +0,124 \hat j )Am^{-1}+(3.82\hat i +1.91 \hat j)Am^{-1}=\\=(3.32 \hat i+2.03 \hat j) Am^{-1}$

Answer: The net magnetic field at the point (25,0) in unit vector notation is $(3.32 \hat i+2.03 \hat j) Am^{-1}$