A straight wire carrying a current $10 A$ is bent into a semicircular arc of radius $5 c m$ . The magnitude of magnetic field at the center is

1.5 \times 10^{- 5} T

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3.14 \times 10^{- 5} T

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6.28 \times 10^{- 5} T

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19.6 \times 10^{- 5} T

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Solution

The correct option is C $6.28 \times 10^{- 5} T$
The magnetic field at the center of semi-circular wire is :
$B = \frac{μ_{0}}{4 π} \times \frac{π i}{r}$

$\Rightarrow B = 10^{- 7} \times \frac{π \times 10}{5 \times 10^{- 2}}$

$= 6.28 \times 10^{- 5} T$

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A straight wire carrying a current 10 A is bent into a semicircular arc of radius 5 cm. The magnitude of magnetic field at the center is

The correct option is C 6.28×10−5TThe magnetic field at the center of semi-circular wire is :B=μ04π×πir⇒B=10−7×π×105×10−2=6.28×10−5 T

A straight wire carrying a current $10 A$ is bent into a semicircular arc of radius $5 c m$ . The magnitude of magnetic field at the center is

The correct option is C $6.28 \times 10^{- 5} T$
The magnetic field at the center of semi-circular wire is :
$B = \frac{μ_{0}}{4 π} \times \frac{π i}{r}$

$\Rightarrow B = 10^{- 7} \times \frac{π \times 10}{5 \times 10^{- 2}}$

$= 6.28 \times 10^{- 5} T$