Question

A long, straight wire carrying a current of 1.0 A is placed horizontally in a uniform magnetic field B = 1.0 × 10−5 T pointing vertically upward figure. Find the magnitude of the resultant magnetic field at the points P and Q, both situated at a distance of 2.0 cm from the wire in the same horizontal plane.
Figure

Answer 1

Explanation:

Step 1:

Given data in the question  :-

Equal magnetic field,$B_{0}=1.0 \times 10^{-5} T$   ( Vertically upwards )

Dividing the point from the wire d = 2 cm = 0.02 m

Step 2:

The intensity of the magnetic field is the

  $B_{o}=\frac{\mu_{0} t}{2 \pi d}$

        $=\frac{4 \pi \times 10^{-7} \times 1}{2 \pi \times 0.02}$

        $=\frac{2 \times 10^{-7} \times 1}{0.02}$

        $=\frac{200 \times 10^{-7}}{2}$

        $=\frac{100 \times 10^{-7}}{2}$

        $=100 \times 10^{-7}$

        $=1 \times 10^{-5} T$

Step 3:

Total magnetic field at point P:

$\begin{array}{c}{B_{P}=B_{W}+B_{o}} \\{B_{P}=1.0 \times 10^{-5} T+1 \times 10^{-5} T}\end{array}$

$B_{P}=10^{-5} T(1+1)$

    $=2 \times 10^{-5} T$

Total magnetic field at point Q :

$B_{Q}=B_{W}-B_{o}$

     $=1.0 \times 10^{-5} T-1 \times 10^{-5} T$

    $=10^{-5} T(1-1)$

    = 0

The magnitude of the resultant magnetic field at the points P and Q are $2\times 10^{-5}$ and 0

Answer 2

Answer:

μ0 = 4π × 10–7T-m/A

r = 2cm = 0.02m,

I = 1A,  

vector B = 1 × 10–5T

We know: Magnetic field due to a long straight wire carrying current ::

= μ0I/2πr

pls preview the 1st attachment for next step calculation....

hope this  helps uhh...!!!!