Question

2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is the magnitude of the magnetic field at a point that is 5.0 cm from each of the wires

Answer 1

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

Where $B =$ magnetic field due to long wires, $\mu_{0} =$ $4\pi \times10^{-7}$, $R =$ perpendicular distance from wire to given point

From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

 $B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

 $B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

 $B = 5.33\times10^{-5} T$

Therefore, the magnitude of magnetic field at given point = $5.33\times10^{-5} T$