A current $I$ flows along a thin wire shaped as a regular polygon with $n$ sides which can be inscribed into a circle of radius $R$ . Find the magnetic induction at the centre of the polygon.

B = \frac{n μ_{0} I tan (\frac{π}{n})}{2 π R}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B = \frac{n μ_{0} I tan (\frac{π}{n})}{π R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B = \frac{n μ_{0} I tan (\frac{2 π}{n})}{π R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B = \frac{n μ_{0} I tan (\frac{2 π}{n})}{2 π R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $B = \frac{n μ_{0} I tan (\frac{π}{n})}{2 π R}$
For polygon we can write

$P = (2 R) sin (\frac{α}{2})$

$P = (2 R) sin (\frac{2 π}{2 n})$

$P = (2 R) sin (\frac{π}{n})$

$P (n) = n (2 R) sin (\frac{π}{n})$

$\frac{P (n)}{n} = 2 R sin (\frac{π}{n})$

Let, $a = \frac{P (n)}{n} = 2 R sin (\frac{π}{n})$
$b = R cos (\frac{π}{n})$

Therefore,
$B_{1} = \frac{μ_{0} I}{4 π b} (cos ϕ_{1} + cos ϕ_{2})$

$B_{1} = \frac{μ_{0} I}{4 π b} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{a}{\sqrt{b^{2} + \frac{a^{2}}{2}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

$B_{1} = \frac{μ_{0} I a}{4 π b} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1}{\sqrt{b^{2} + \frac{a^{2}}{2}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

Magnetic induction due to $n -$ sides

$B = n B_{1}$

Using above values, $\frac{1}{\sqrt{b^{2} + \frac{a^{2}}{2}}} = 1$

$B = n [\frac{μ_{0} I}{4 π}] ⎡ ⎢ ⎢ ⎣ \frac{2 R sin (\frac{π}{n})}{R cos (\frac{π}{n})} ⎤ ⎥ ⎥ ⎦$
$B = [\frac{n μ_{0} I}{2 π R}] [tan (\frac{π}{n})]$

Suggest Corrections

Similar questions

i

n

R

I

n

R

n \to \infty

i

n

R

2 π r

i

B

Join BYJU'S Learning Program