VITEEE 2015 Paper

For a dipole at position (R, Q)
BR =

µ0

4π

.

2Mcosθ

R3

... (i)
and BQ =

µ0

4π

.

Mcosθ

R3

... (ii)
Also tan Φ =

BV

BH

= -

BR

BQ

... (iii)

Dividing eq. (i) by (ii)

BR

BQ

=

2cosθ

sinθ

= cot θ ... (iv)
From eq. (iii) and (iv)
tan Φ = - 2 cot Φ
From figure , θ = 90° + λ
∴ tan ϕ = - 2 cot (90 + λ)
tan ϕ = 2 tan λ