Question

A wire carrying a current of 2.85 A has a straight segment of 2.35 m. Determine the magnitude of the magnetic field due to this segment of wire at a point P which is at the midpoint of the straight segment of wire and a distance of 0.275 m perpendicular to the wire.

Answer 1

Answer:

$2.02\times 10^{-6} T$

Explanation:

We are given that

Current,I=2.85 A

Length of segment=l=2.35 m

d=0.275 m

We have to find the magnitude of magnetic field due to this segment of wire at a point P which is at midpoint of the straight wire.

$r=\frac{l}{2}=\frac{2.35}{2}=1.175 m$

$\theta=tan^{-1}(\frac{r}{d})=tan^{-1}(\frac{1.175}{0.275})=76.82^{\circ}$

$\theta=\theta_1=\theta_2$

Magnetic field,B=$\frac{\mu_0I}{4\pi d}(sin\theta_1+sin\theta_2)=\frac{\mu_0I}{4\pi d}(2sin\theta)$

Where $\frac{\mu_0}{4\pi}=10^{-7}$

Using the formula

$B=\frac{10^{-7}\times 2.85}{0.275}(2sin(76.82))=2.02\times 10^{-6} T$