Magnetic Effects & Magnetism

JEE Main level, single correct, medium

43 of 276

Question 43 of 276

As shown in the figure, a long straight conductor with semicircular arc of radius π10m\dfrac{\pi }{10}m is carrying current I=3AI=3A. The magnitude of the magnetic field. at the center OO of the arc is: (The permeability of the vacuum =4π×10−7NA−2=4\pi \times {10}^{-7}{NA}^{-2} )

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Answer: (D) 3μT3\mu T

The formula to calculate the magnetic field (BC)\left({B}_{C}\right) produced at the centre of a circular current carrying conductor is given by
BC=μoI4πrθ.......................(1){B}_{C}=\dfrac{{\mu }_{o}I}{4\pi r}\theta .......................(1)
where, μ0{\mu }_{0} is the permeability of free space, II is the current through the conductor, rr is the radius of the circular conductor and θ\theta is the angle subtended by the arc at the centre.
From the given figure, it is clear that the arc of the conductor makes an angle π\pi at its centre.
Substitute the known values of the parameters into equation (1) to calculate the required magnetic field.
BC=4π×10−7NA−2×3A4×π×π10m×π=3×10−6T×106μT1T=3μT\begin{matrix}{B}_{C} & = & \dfrac{4\pi \times {10}^{-7}N{A}^{-2}\times 3A}{4\times \pi \times \dfrac{\pi }{10}m}\times \pi \\ & = & 3\times {10}^{-6}T\times \dfrac{{10}^{6}\mu T}{1T} \\ & = & 3\mu T\end{matrix}

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