A spiral of two turns with an inner radius $R$ and outer radius $2 R$ is placed as shown in the figure. Find the magnetic field at the centre of the spiral.

\frac{μ_{0} I}{R} ln 2

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\frac{μ_{0}}{R} ln 3

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\frac{μ_{0} i}{R}

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\frac{μ_{0} i}{4 π} ln 3

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Solution

The correct option is A $\frac{μ_{0} I}{R} ln 2$
$\begin{matrix} B = \int_{a}^{b} \frac{μ_{0} N T}{2 (b - a)} \frac{d r}{r} = \frac{μ_{0} N t}{2 (b - a)} l n (\frac{b}{a}) = \frac{μ_{0} \times 2 I}{2 (2 R - R)} l n (\frac{2 R}{R}) = \frac{μ_{0} I}{R} l n 2 \end{matrix}$

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A spiral of two turns with an inner radius R and outer radius 2R is placed as shown in the figure. Find the magnetic field at the centre of the spiral.

The correct option is A μ0IRln2B=∫baμ0NT2(b−a)drr=μ0Nt2(b−a)ln(ba)=μ0×2I2(2R−R)ln(2RR)=μ0IRln2

A spiral of two turns with an inner radius $R$ and outer radius $2 R$ is placed as shown in the figure. Find the magnetic field at the centre of the spiral.

The correct option is A $\frac{μ_{0} I}{R} ln 2$
$\begin{matrix} B = \int_{a}^{b} \frac{μ_{0} N T}{2 (b - a)} \frac{d r}{r} = \frac{μ_{0} N t}{2 (b - a)} l n (\frac{b}{a}) = \frac{μ_{0} \times 2 I}{2 (2 R - R)} l n (\frac{2 R}{R}) = \frac{μ_{0} I}{R} l n 2 \end{matrix}$