Question

The magnitude of magnetic field produced by an infinitely long current carrying conductor at a perpendicular
distance R from the conductor is given by B = Hol and its direction is given by right hand thumb rule, where
2R
Ho = Permeability of vacuum (a constant) = 4 TT * 10-7 Tm/A
B = Magnetic field strength (in tesla)
i = Current flowing in conductor (in ampere)
R = Distance from conductor in meter)
Then consider a current 10A is flowing in a straight conductor of infinite length. At what
distance in meters from the wire the magnetic field will be 5 x 10-7 T ?​

Answer 1

Answer:  

Well you answer is 0.4 Meters. Just substitute in the formula to find "R"

Explanation:

You are an Allen student, aren't you?

Since all values are given in their base units, no conversion is required.

Answer 2

Given:

The magnitude of magnetic field produced by an infinitely long current carrying conductor at a perpendicular distance R from the conductor is given by:

$B = \dfrac{ \mu_{0}}{4\pi} \bigg ( \dfrac{2i}{R} \bigg)$

To find:

Distance at which the magnetic field intensity will be 5 × 10^(-7) Tesla ?

Calculation:

As per the question, the magnetic field intensity at a distance R from the centre of a infinitely long current carrying wire (R > radius of wire) is given as :

$\therefore \: B = \dfrac{ \mu_{0}}{4\pi} \bigg ( \dfrac{2i}{R} \bigg)$

Now, putting all the values in SI unit;

$\implies \: 5 \times {10}^{ - 7} = {10}^{ - 7} \times \bigg ( \dfrac{2 \times 10}{R} \bigg)$

$\implies \: 5 = \dfrac{2 \times 10}{R}$

$\implies \: 5 = \dfrac{20}{R}$

$\implies \:R = \dfrac{20}{5}$

$\implies \:R = 4 \: m$

So, the required Magnetic Field Intensity will be obtained at R = 4 m from the centre of the wire.