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AIIMS 2016
Exam
AIIMS
Year
2016
Subject
Physics
Chapter
Magnetic Effects of Current

AIIMS 2016 · Physics · Magnetic Effects of Current

The magnetic field at the centre O of the arc shown in the figure is

The magnetic field at the centre OO of the arc shown in the figure is The magnetic field at the centre O of the arc shown in the figure is

Options

  1. A2I(2+π)×10−7r2 I(\sqrt{2}+\pi) \times \frac{10^{-7}}{r}
  2. B2I(2+π4)×10−7r2 I\left(\sqrt{2}+\frac{\pi}{4}\right) \times \frac{10^{-7}}{r}
  3. CI(2+π)×10−7rI(\sqrt{2}+\pi) \times \frac{10^{-7}}{r}
  4. DI(2+π4)×10−7rI\left(\sqrt{2}+\frac{\pi}{4}\right) \times \frac{10^{-7}}{r}

Correct answer

B. 2I(2+π4)×10−7r2 I\left(\sqrt{2}+\frac{\pi}{4}\right) \times \frac{10^{-7}}{r}

Step-by-step solution

The magnetic field at the centre O of the arc shown in the figure is
Here, a=r2a=\frac{r}{\sqrt{2}} Magnetic field at point OO due to ABA B is B1=μ04πIa=μ0I4π(r/2)B_1=\frac{\mu_0}{4 \pi} \frac{I}{a}=\frac{\mu_0 I}{4 \pi(r / \sqrt{2})} Magnetic field at point OO due to BCDB C D is B2=μ0I4πrB_2=\frac{\mu_0 I}{4 \pi r}\left(\frac{\pi}{2}\right)MagneticfieldatpointMagnetic field at pointOduetodue toD EisisB_3=\frac{\mu_0 I}{4 \pi a}=\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}ResultantmagneticfieldatpointResultant magnetic field at pointOisisB=B1+B2+B3=μ0I4π(r/2)+μ0I4πr(π2)+μ0I4π(r/2)=μ0I4πr(2+π2+2)=μ0I4πr(22+π2)=μ0II4πr(2+π4)=10−7×2Ir(2+π4)[∵μ04π=10−7]\begin{aligned} B & =B_1+B_2+B_3 \\ & =\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}+\frac{\mu_0 I}{4 \pi r}\left(\frac{\pi}{2}\right)+\frac{\mu_0 I}{4 \pi(r / \sqrt{2})} \\ & =\frac{\mu_0 I}{4 \pi r}\left(\sqrt{2}+\frac{\pi}{2}+\sqrt{2}\right)=\frac{\mu_0 I}{4 \pi r}\left(2 \sqrt{2}+\frac{\pi}{2}\right) \\ & =\frac{\mu_0 I I}{4 \pi r}\left(\sqrt{2}+\frac{\pi}{4}\right) \\ & =\frac{10^{-7} \times 2 I}{r}\left(\sqrt{2}+\frac{\pi}{4}\right) \quad\left[\because \frac{\mu_0}{4 \pi}=10^{-7}\right]\end{aligned}$
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