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BITSAT 2017
Exam
BITSAT
Year
2017
Subject
Physics
Chapter
Magnetic Effects of Current

BITSAT 2017 · Physics · Magnetic Effects of Current

What is the magnetic field at the centre of arc in the figure below?

What is the magnetic field at the centre of arc in the figure below?

What is the magnetic field at the centre of arc in the figure below?

Options

  1. Aμ04π⋅2Ir[2+π]\dfrac{{\mu}_{0}}{4\pi} \cdot \dfrac{2I}{r}\left[\sqrt{2} + \pi\right]
  2. Bμ04π⋅2Ir[2+π4]\dfrac{{\mu}_{0}}{4\pi} \cdot \dfrac{2I}{r}\left[\sqrt{2} + \dfrac{\pi}{4}\right]
  3. Cμ04π⋅Ir[2+π]\dfrac{{\mu}_{0}}{4\pi} \cdot \dfrac{I}{r}\left[\sqrt{2} + \pi\right]
  4. Dμ04π⋅Ir[2+π4]\dfrac{{\mu}_{0}}{4\pi} \cdot \dfrac{I}{r}\left[\sqrt{2} + \dfrac{\pi}{4}\right]

Correct answer

B. μ04π⋅2Ir[2+π4]\dfrac{{\mu}_{0}}{4\pi} \cdot \dfrac{2I}{r}\left[\sqrt{2} + \dfrac{\pi}{4}\right]

Step-by-step solution

Distance of straight path from OO

=r2= \dfrac{r}{\sqrt{2}}

Hence, B=B1+B2B = {B}_{1} + {B}_{2}

=μ04π⋅2Ir2+μ04π2\piIr(14)= \dfrac{{\mu}_{0}}{4\pi} \cdot \dfrac{2I}{\dfrac{r}{\sqrt{2}}} + \dfrac{{\mu}_{0}}{4\pi}\dfrac{2\piI}{r}\left(\dfrac{1}{4}\right)

=μ04π⋅2Ir[2+π4]= \dfrac{{\mu}_{0}}{4\pi} \cdot \dfrac{2I}{r}\left[\sqrt{2} + \dfrac{\pi}{4}\right]

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