Question

A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of 4.0 × 10−4 T parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire.

Answer 1

A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of 4.0 × 10−4 T parallel to the current. then the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire is $5 \times 10^{-4} T$

Explanation:

Step 1:

Given data in the question  :

Same magnetic field  $B_{0}=4.0 \times 10^{-4} T$

Current magnitude , I = 30 A

Divide the point from wire, d = 0.02 m

Step 2:

Therefore, the magnetic field is given by current in the wire

$B=\frac{\mu_{0} i}{2 \pi d}$

 $=\frac{2 \times 10^{-7} \times 30}{0.02}$

 $=\frac{60 \times 10^{-7}}{0.02}$

 $=\frac{6000 \times 10^{-7}}{2}$

 $=3000 \times 10^{-7}$

$=3 \times 10^{-4} T$

$B_0$ is at right angles to B.

Step 3:

The magnetic field which results         

$=\sqrt{16 \times 10^{-8}+9 \times 10^{-8}}$

$=\sqrt{10^{-8} \times(16+9)}$

$=\sqrt{10^{-8} \times(25)}$

$=5 \times 10^{-4} T$