VITEEE 2015 Paper

Due to current through side AB Magnetic field at the centre O
B1 =

µ0I

4πa

[sinθ1+sinθ2]

As the magnetic field due to each of the 24. For a dipole at position (R, Q) three sides is the same in magnitude and direction.
∴ Total magnetic field at O is sum of all the fields.
i.e. B = 3B1 =

3µ0I

4πa

[sinθ1+sinθ2]
Here, tan θ1 =

AD

OD

⇒ tan 60° =

l 2

a

⇒ a =

l

2√3

=

9×10−2

2√3

Now B
= 3 ×

4π×10−7×2

4π× 9×10−2 2√3

[sin 60° + sin 60°]
=

4√3

9

× 10−5 [

√3

2

+

√3

2

]
1.33 x 10−5 T