# Three long current-carrying wires are perpendicular to the xy-plane passing through the points (x,y)= (0,-1 cm), (0, +1 cm) and (+1 cm,0) respectively. If each wire carries a current I=2 A in the +z-direction, what is the magnetic field at the origin?

## I have to apply Biot-Savart Law which is $B = \setminus \frac{\setminus {\mu}_{o} I}{4 \setminus \pi} \setminus \int \setminus \frac{\mathrm{ds} \setminus \times r}{{r}^{2}}$ and I am not sure how use it in this problem. The right answer is $\left(0 , - 4.0 \setminus \times {10}^{-} 5 , 0\right) T$.

Jun 29, 2016

see below

#### Explanation:

i think Bio Savart is OTT for this

you just need to look at the symmetry. You will see in the diagram, using the Right Hand Thumb Rule, that the fields due to wires a and b cancel out at the origin. Because they are equal in magnitude and oppsite in direction, their superimposition equals zero.

That leaves us with the field due to c, which at the origin, points down the y axis.

it's magnitude will be $\textcolor{red}{B = \frac{{\mu}_{o} I}{2 \pi r}} = \frac{4 \pi \cdot {10}^{- 7} \times 2}{2 \pi 0.01} = 4 \cdot {10}^{- 5} T$

which is how we get $\left(0 , - 4.0 \setminus \times {10}^{-} 5 , 0\right) T$

[ the formula in red can be derived using Bio Savart, or more easily from Ampere's Law]