Question 40 of 276

A straight wire carrying a current of $14A$ is bent into a semicircular arc of radius $2.2cm$ as shown in the figure. The magnetic field produced by the current at the centre $O$ of the arc is ____ $\times {10}^{–4}T$

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Answer: 2

It is given that
$i=14A R=2.2\times {10}^{-2}m$
The magnetic field at the centre $O$ is
$B=\dfrac{{\mu }_{0}i}{4R} =\dfrac{4\pi \times {10}^{-7}\times 14}{4\times 2.2\times {10}^{-2}} =2\times {10}^{-4}T$