What is the magnetic field at the cemter of a coil in the form of a square of side` 4sqrt2 cm` and carrying a current of `4 A` ?

Options

A. `8 xx 10^(-7)T`
B. `6 xx 10^(-7)T`
C. `4 xx 10^(-7)T`
D. `3 xx 10^(-7)T`

Answer

A

Explanation

Magnetic field at O due to the current carrying conductor CD is `B_I=(mu_0)/(4pi).1/r(sin phi_1+sinphi_2)` where`I=4 A, r=2sqrt2cm,phi_1=phi_2=45 and sin 45=1/sqrt2 thereforeB_1=10^(-7)xx4/(2sqrt2)=2 xx 10^(-7)T therefore ` Total magnetic field at `O=4 B_1=8 xx 10^(-7)T`.