Question 9 of 276

A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. Find the magnetic field at centre, when a current $I$ passes through coil :

A$\dfrac{{\mu }_{0}I}{8}\left[\dfrac{a+b}{a-b}\right]$B$\dfrac{{\mu }_{0}I}{4(a-b)}\left[\dfrac{1}{a}-\dfrac{1}{b}\right]$C$\dfrac{{\mu }_{0}I}{8}\left(\dfrac{a-b}{a+b}\right)$D$\dfrac{{\mu }_{0}IN}{2(b-a)}{\log }_{e}\left(\dfrac{b}{a}\right)$

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Answer: (D) $\dfrac{{\mu }_{0}IN}{2(b-a)}{\log }_{e}\left(\dfrac{b}{a}\right)$

Recall the formula of the magnitude of the magnetic field in terms of current and permeability in free space, $B={\int }_{a}^{b}\dfrac{{\mu }_{0}dNi}{2x}={\int }_{a}^{b}\dfrac{{\mu }_{0}\left(\dfrac{N}{b-a}dx\right)i}{2x}=\dfrac{{\mu }_{0}Ni}{2(b-a)}ln\left(\dfrac{b}{a}\right)$.

JEE Main Physics practice question on Magnetic Effects & Magnetism (Biot-Savart's Law and Ampere's Circuital Law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.