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A very long wire ABDMNDC is shown in fig...

A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is :

A

`(mu_0I)/(2piR)(pi+1/sqrt2)`

B

`(mu_0I)/(2R)`

C

`(mu_0I)/(2piR)(pi-1/sqrt2)`

D

`(mu_0I)/(2piR)(pi+1)`

Text Solution

Verified by Experts

The correct Answer is:
A

`vecB_0=(vecB_0)_A+(vecB_0)_B+(vecB_0)_C`
`(mu_0I)/(4piR)[sin90^"@-sin45^@]ox+(mu_0I)/(2R) o.+(mu_0I)/(4piR)(sin45^@ - sin90^@)o.`
`(mu_0I)/(4piR)[1-1/sqrt2]+(mu_0I)/(2R)+(mu_0I)/(4piR)[1/sqrt2+1]o.`
`(mu_0I)/(4piR)[-1+1/sqrt2+2pi+1/sqrt2+1]o.=(mu_0I)/(4piR)[sqrt2+2pi]o.(mu_0I)/(2piR)[1/sqrt2+pi]o.`
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