Question

Magnetic field strength at the centre of regular pentagon made of a conducting wire of uniform cross section area as shown in figure is : (i amount of current enters at A and leaves at E)

• A. $\frac{5{\mu }_{0}i}{4\pi a}\left[2sin\frac{{72}^0}{2}\right]$
• B. 0
• C. $\frac{3{\mu }_{0}i}{4\pi a}\left[2sin\frac{{72}^0}{2}\right]$
• D. $\frac{{\mu }_{0}i}{4\pi a}\left[2sin\frac{{72}^0}{2}\right]$

Answer 1

The correct option is A $\frac{5{\mu }_{0}i}{4\pi a}\left[2sin\frac{{72}^0}{2}\right]$

$InthisthemagneticfieldduetoeachstraightconductorisbeingaddedandisintheinwarddirectionthemagneticfieldduetoonestraightconductorB=\frac{{\mu }_{0}i(sin\alpha +sin\beta )}{4\pi r}Hereboth\alpha =\beta ={72/2}^{0}degreesNowputtingvalueB=\frac{{\mu }_{0}i\left(2sin{72/2}^0\right)}{4\pi r}Nowtotalmagneticfieldatthecentreis4B=4\frac{{\mu }_{0}i\left(2sin{72/2}^0\right)}{4\pi r}whereris\frac{4a}{5}thereforetotalmagneticfieldbecomes=\frac{5{\mu }_{0}i\left(2sin{72/2}^0\right)}{4\pi a}thereforeoptionA$