Question

The diagram below shows a long wire with a right angle bend. A current of I = 2.78 A flows through the wire. Determine the magnetic field (in unit vector notation) at the point P which is a distance a = 1.96 cm from the corner of the bend. (Assume that the +xaxis is directed to the right, the +y axis is directed up, and the +2 axis is toward you.) 

Answer 1

Solution:

Current, I= 2.78 A

Distance from the corner of the bend, a =1.96cm

Magnetic Field, $B=\frac{\mu _{0}I}{4\pi a}(sin\alpha +sin\beta )$

Since the magnetic field due to horizontal wire is zero.

$B=\frac{\mu _{0}I}{4\pi a}$

$\mu _{0}=4\pi \times 10^{-7}H/m$

$B=\frac{4\pi \times 10^{-7}H/m\times 2.78A}{4\pi \times (1.96\times 10^{-2}m)}$

$B=14.18\times 10^{-6}T$