Question

A current of 5a is flowing from south to north in a straight wire. find the magnetic field due to 1 cm piece of wire at a point 1 m north east from the piece of wire

Answer 1

Final Answer : 3.54 x. 10^(-9) Tesla

Steps:
1) We know the Magnetic Field due to small current carrying element.
Since, Distance from wire >>> length of wire.
Therefore, angle subtended at wire due to far away point is almost same as taken from horizontal .

2) Angle subtended will be,
$\theta \: = 45 \degree$
Length of wire (segment of wire) , L = 0.01 m
Distance of Point from centre of wire ,r = 1 m
Current in wire, I = 5 A

Using relation From biot Savart Law, as shown in pic .
Magnetic Field at that point is
$3.54 \times {10}^{ - 9} tesla$

Direction of Magnetic Field will be $\boxed { Into \:the \:Page }$

Answer 2

Answer: The magnetic field of the wire is $dB = 3.5\times10^{-9} T$

Explanation:

Given that,

Current i = 5 A

Length l = 1 cm

Radius r = 1 m

Angle $\theta = 45^{0}$

We know that,

The magnetic field is

$dB = \dfrac{\mu_{0}il}{r^{2}}sin\theta$.....(I)

Put the all values in equation (I)

$dB = 10^{-7}\times \dfrac{5\times 0.01}{1}\times \dfrac{1}{\sqrt{2}}$

$dB = 3.5\times10^{-9} T$

Hence, the magnetic field of the wire is $dB = 3.5\times10^{-9} T$