Question

A straight wire of length 0.62 m carries a conventional current of 0.8 A. What is the magnitude of the magnetic field made by the current at a location that is a perpendicular distance 2.9cm from the center of the wire? Use both the exact equation and the approximate equation to calculate the field.

Short answer

The magnetic field for approximate is,$5.52\times {10}^{-6}\mathrm{T}$ and the magnetic field for exact is, $5.49\times {10}^{-6}\mathrm{T}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

· The length was of the wire is, $L=0.62m$.

· The conventional current is,$I=0.8A$ .

· The perpendicular distance of wire is $r=2.9cm$.

Significance of the magnetic field

The area in which the force of magnetism acts around a magnetic material or a moving electric charge is known as the magnetic field.

Determination of the magnetic field

The expression of magnetic field for approximate is expressed as,

$B=\frac{{\mu }_{0}l}{2\pi r}$ …(1)

Here I is the conventional current and r is the perpendicular distance of the wire.

Substitute 0.8 A for I, $2.9\times {10}^{-2}m$for r, and $4\pi \times {10}^{-7}H/m$for ${\mu }_0$in the equation (1).

$$\begin{gathered} B=\frac{\left(4\pi \times {10}^{-7}H/m\right)\times 0.8A}{2\pi \times 2.9\times {10}^{-2}m} \\ =5.52\times {10}^{-6}T \end{gathered}$$

Hence the magnetic field for approximate is, $5.52\times {10}^{-6}\mathrm{T}$.

The expression of magnetic field for exact is expressed as,

$B=\frac{{\mu }_{0}IL}{2\pi r}\times \frac{1}{\sqrt{4r^2+L^2}}$ …(2)

Here L is the length of wire.

Substitute 0.8 A for I, $2.0\times {10}^{-2}m$for r 0.62m for L and $4\pi \times {10}^{-7}H/m$for u in the equation (2).

role="math" localid="1668547379159" $B=\frac{{\mu }_{0}IL}{2\pi r}\times \frac{1}{\sqrt[]{4f^2+L^2}}$

Hence the magnetic field for exact is,$5.52\times {10}^{-6}$