Question

Two long parallel wires 6.50cm apart carry 18.5A currents in the same direction. Determine the magnetic field strength at a point 12.0cm from one wire and 13.6cm from the other.

Answer 1

Answer:

$5.8\cdot 10^{-5}T$

Explanation:

The magnetic field produced by a current-carrying wire is:

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the field is given by the right-hand rule: the thum is placed in the same direction of the current, and the other fingers "wrapped" around the thumb gives the direction of the field.

First of all, here we are finding the magnetic field strength at a point 12.0cm from one wire and 13.6cm from the other: since the two wires are 6.50 cm apart, this means that the point at which we are calculating the field is located either on the left or on the right of both wires. So, by using the right-hand rule for both, we see that both fields go into the same direction (because the currents in the two wires have same direction).

Therefore, the net magnetic field will be the sum of the two magnetic fields.

For wire 1, we have:

$I_1=18.5 A\\r_1=12.0 cm =0.12 m$

So the field is

$B_1=\frac{(4\pi \cdot 10^{-7})(18.5)}{2\pi(0.12)}=3.1\cdot 10^{-5} T$

For wire 2, we have:

$I_2=18.5 A\\r_2=13.6 cm=0.136 m$

So the  field is

$B_2=\frac{(4\pi \cdot 10^{-7})(18.5)}{2\pi(0.136)}=2.7\cdot 10^{-5} T$

Therefore, the total magnetic field is

$B=B_1+B_2=3.1\cdot 10^{-5}+2.7\cdot 10^{-5}=5.8\cdot 10^{-5}T$