Infinite number of straight wires each carrying current I are equally placed as shown in the figure Adjacent wires have current in opposite direction Net magnetic field at point `P` is
.

Options

A. `(mu_(0)I)/(4pi)(In2)/(sqrt3a)hatk`
B. `(mu_(0)I)/(4pi)(In4)/(sqrt3a)hatk`
C. `(mu_(0)I)/(4pi)(In4)/(sqrt3a)(-hatk)`
D. zero

Answer

B

Explanation

`B_(1)=(mu_(0))/(4pi) (I)/(a) (sinalpha +sinbeta) =(mu_(0))/(4pi) (Ixx2 sin 30^(@))/(a cos 30^(@))`
`oversetrarr(B) = B_(1) hatk - B_(2)hatk +B_(3)hatk -B_(4)hatk+......`
`= (mu_0)/(4pi)(Ixx2 sin 30^@)/(a cos 30^(@))hatk - (mu_0)/(4pi) (I xx 2 sin 30^(@))/(2a cos 30^(@))hatk + `
`(mu_(0))/(4pi) (I xx 2 sin30^(@))/(3a cos 30^(@))hatk+......`
`=(mu_(0))/(4pi)(2I)/(sqrt3a) hatk - (mu_0)/(4pi) (2I)/(sqrt3 2a) hatk + mu_(0)/(4pi)(2I)/(sqrt33a) hatk+....`
` =(mu_0)/(4pi) (2I)/(sqrt3a) [1 - (1)/(2) + (1)/(3) - (1)/(4) + .... ]hatk`
`=(mu_0)/(4pi) (2I)/(sqrt3a) log_(e) (1+1)hatk=(mu_(0)I)/(4pi sqrt3a )log_(e) 4hatk` .