A current path shaped as shown in figure produces a magnetic field at P, the centre of the arc. If the arc subtends an angle of 30° and the radius of the arc is 0.6 m. What is the magnitude of the field at P, if the current is 3.0 A?

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$2.62 \times 10^{- 6} T$

$2.62 \times 10^{- 7} T$

$2.62 \times 10^{- 8} T$

$3.62 \times 10^{- 7} T$

answer is B.

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Detailed Solution

Magnetic field at centre of the circular loop,

$B = \frac{μ_{0} Ni}{2 R}$

Magnetic field due to an arc of a circle at the centre,

$B = (\frac{θ}{2 π}) \frac{μ_{0} i}{2 R} = (\frac{μ_{0}}{4 π}) (\frac{i}{R}) θ$

Here, $θ = 30^{\circ} and i = 3 A, R = 0 .6 m$

$\begin{matrix} B = (\frac{μ_{0}}{4 π}) (\frac{3}{0 .6}) (\frac{π}{6}) \\ = \frac{10^{- 7} \times 3 \times π}{0 .6 \times 6} = 2 .62 \times 10^{- 7} T \end{matrix}$

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