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A current path shaped as shown in the 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 are the magnitude and direction of the field produced at $P$ if the current is $3.0 A$ .
current path

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

$B = 2.062 \times 10^{- 7} T$

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

$B = 2.602 \times 10^{- 7} T$

answer is A.

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Detailed Solution

The magnetic field at $P$ due to the straight segments $A C$ and $D E$ is zero. $C D$ is arc of circle.

$∴ B = (\frac{θ}{2 π}) (\frac{μ_{0} i}{2 R}) B = (\frac{μ_{0}}{4 π}) (\frac{i}{R}) θ B = (10^{- 7}) (\frac{3.0}{0.6}) (\frac{π}{6}) B = 2.62 \times 10^{- 7} T$

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