Question

A straight wire carrying current of 12a is bent in to a semicircular arc of radius 2 cm as shown in fig.a c onsider the magnetic field b at the centre of the arc

Answer 1

Answer: The magnetic field at the center of the arc is $B = 1.88\times10^{-4}T$

Explanation:

Given that,

Current = 12 A

Radius of arc = 2 cm

We know that,

The magnetic field at the center of the arc is

$B = \dfrac{\mu_{0} I}{4r}$

$B = \dfrac{4\pi\times10^{-7}\times12 A}{4\times2\times10^{-2}m}$

$B = 6\pi\times10^{-5} T$

$B = 1.88\times10^{-4}T$

Hence, the magnetic field at the center of the arc is $B = 1.88\times10^{-4}T$

Answer 2

Answer:

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Explanation:

1.9 Gauss, inwards and upwards