Necker Cube

For (I), the four currents contribute magnetic field vectors with directions: $i+j,i+k,i-j,i-k$ which sum up to $i$ .

For (II), the 6 currents contribute magnetic field vectors with directions: $i+j,j-k,i-k,i+j,j-k,i-k$ which sum up to $4(i+j-k)$ . So we have $cos\theta = {(i+j-k) . (i) \over \sqrt{3}\sqrt{1}}$ . So $\theta = acos({1 \over \sqrt{3}}) = 54.73^o$ .

$\vec {B_{I}}$ = $\hat {x}$

$\vec {B_{II}}$ = $\hat {x} + \hat {y} + \hat{z}$

$\theta = cos^{-1} \frac{\vec {B_{I}}.\vec {B_{II}}}{|B_{I}| |B_{II}}|$ = $cos^{-1}\frac{1}{\sqrt3}$ = $54.74^ \circ$

In figure I, the net magnetic field due to current is along +ve $x-axis$ so let $B_{I}=B_{0} \hat{i}$

As, The magnitude of current is same in second case as well so $B_{II}=B_{0} \hat{i} + B_{0} \hat{-j} + B_{0} \hat{k}$ Now using scalar product of vectors will fetch us $\theta = cos^{-1} \frac{1}{\sqrt{3}}$ so, $\theta \approx 54.74$

I decomposed in pairs of vectors and used the right hand rule to determine the direction of the resulting magnetic field fields. I assumed that the resulting magnitude of each resulting vector would be 1 on the resulting axis. So, if I did, y cross - z => 1 unit on x, and so on.

On (I), I ended up with B1 = (2,0,0) On (II), I ended up with B2 = (1,-1,1)

Then I did, Cos(x) = (B1 dot B2)/|B1||B2| Cos(x) = 2/ (2 * (3)^1/2)

Cos(x) = (3^1/2)/3

Then solved for x, using acos((3^1/2)/3) = 54.74