Question

The magnetic field intensity due to a thin wire carrying current I shown in figure, is

A. `(mu_0I)/(2piR)[pi-alpha+tanalpha]`
B. `(mu_0I)/(2piR)[pi-alpha]`
C. `(mu_0I)/(2piR)[pi+alpha]`
D. `(mu_0I)/(2piR)[pi+alpha-tanalpha]`

Answer 1

Correct Answer - A

Magnetic field due to current in wire BA,
`B_(BA)=(mu_0I)/(4pi(OC))[sin alpha+sin alpha]`
`=(mu_0I)/(4piRcos alpha)(2 sin alpha)` (as `OC=Rcos alpha`)
or `B_(BA)=(mu_0I)/(2piR)tanalpha`
It is acting vertically downwards.
`B_(ADB)=[(mu_0I)/(4piR)(2pi-2alpha)]=(mu_0I)/(2piR)(pi-alpha)`
It is acting vertically downwards,
Total magnetic field, `B=B_(BA)+B_(ADB)`
`=(mu_0I)/(2piR)(pi-alpha+tan alpha)`