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  • A straight wire carrying a current of 14 A is bent into a semicircular arc of radius 2.2 cm as shown in the figure. The magnetic field produced by the current at the centre (O) of the arc. i

A straight wire carrying a current of 14 A is bent into a semicircular arc of radius 2.2 cm as shown in the figure.
The magnetic field produced by the current at the centre (O) of the arc. is ________\times 10^{7}T

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

$$ \begin{aligned} & \mathrm{B}_{\text {total }}=\mathrm{B}_1+\mathrm{B}_{\mathrm{II}}+\mathrm{B}_{\mathrm{III}} \\ & \mathrm{B}_{\mathrm{I}}=0 \\ & \mathrm{~B}_{\mathrm{III}}=0 \end{aligned}

Because  \overrightarrow{d l} \times \vec{r}=0

Now, magnetic field due to semicirclaur ring at its center is given by

\begin{aligned} & \mathrm{B}_{II}=\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}} \\ & =\frac{4 \pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}} \\ & =\frac{22}{7} \times \frac{10^{-7} \times 14}{22 \times 10^{-3}} \\ & =2 \times 10^{-7} \\ & =2 \end{aligned}

Posted by

HARSH KANKARIA

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