Magnetic field

Electricity and Magnetism Level 2

The segment of wire shown carries a current i = 5 A i=5 A , where the radius of the circular arc is R = 3 c m R= 3 cm . Determine the magnitude and direction of the magnetic field at the point O.

Assumptions:-

  • μ 0 4 π = 1 0 − 7 \frac{\mu_0}{4\pi}=10^{-7}

  • μ 0 \mu_0 is different from μ \mu .

  • μ = 1 0 − 6 \mu=10^{-6}

26.2 μ T 26.2\mu T outside the screen. 22.2 μ T 22.2\mu T outside the screen. 26.2 μ T 26.2\mu T inside the screen. 22.2 μ T 22.2\mu T inside the screen.

Sahil Silare by Sahil Silare , 20, India

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1 solution

Aryaman Maithani
Jun 8, 2018

The current through the wires do not contribute to the magnetic field as the point lies on both the lines (upon extending).

By Biot-Savart Law , the magnetic field due to a current carrying element is given by:

d B ⃗ = μ 0 i d l ⃗ × r ^ 4 π r 2 d\vec{B} = \frac{\mu_0\text{ } i \text{ } d\vec{l}\times\hat{r} }{4\pi r^2}

It can be seen the the dircetion of d l ⃗ × r ^ d\vec{l}\times\hat{r} has constant direction and can therefore be simplified as:

∴ B ⃗ = μ 0 i 4 π r 2 ∫ d l \therefore \vec{B} = \frac{\mu_0\text{ } i \text{ }}{4\pi r^2} \int dl

∫ d l \int dl is simply the total length of the arc, that is, π r 2 \frac{\pi r}{2}

∴ B ⃗ = μ 0 i 4 π r 2 ( π r 2 ) \therefore \vec{B} = \frac{\mu_0\text{ } i \text{ }}{4\pi r^2} \Big(\frac{\pi r}{2}\Big)

Substituting the values, the magnitude of the field can be calculated and the direction can be determined by using the right hand rule.

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