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Shown in Fig. is a conductor carrying cu...

Shown in Fig. is a conductor carrying current I. Find the magnetic field intensity at point O.

Text Solution

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The magnetic field at the centre of an arc is equal to
`B=(mu_0I)/(4pir) theta`
Magnetic field due to arc (1), `B_1=(mu_0I theta)/(4pixx3r)(-hatk)`
Magnetic field due to arc (2), `B_2=(mu_0I theta)/(4pixx2r)(hatk)`
Magnetic field due to arc (3), `B_1=(mu_0I theta)/(4pir)(-hatk)`

Net magnetic field, `B=B_1+B_2+B_3`
Hence, net `B=(mu_0I)/(4pi)[-1/r+1/(2r)-1/(3r)] theta (hatk)=-(5mu_0I theta)/(24 pir) hatk`
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