If a long horizontal conductor is bent as shown in figure and a current I is passed in it, find the magnitude and direction of magnetic field induction at the centre of circular part.
A
`(mu_0I)/(2pi r)(pi-1)`
B
`(mu_0I)/(2pi r)(pi+1)`
C
`(mu_0I)/(2pi r)(2pi-1)`
D
`(mu_0I)/(2pi r)(2pi+1)`
Text Solution
Verified by Experts
The correct Answer is:
A
The magnetic field due to straight portion of the conductor is `B_1=(mu_0I)/(2pir)` (upward) The magnetic induction due to circular loop is `B_2=(mu_0I)/(2pi r)` (upward) Net magnetic induction is givent by , ` B = B_2 -B_1=(mu_0I)/(2r)-(mu_0I)/(2pir)` `=(mu_0 I)/(2 pi r)(pi -1) ` downward .
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