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Consider a straight conductor XY carrying current I. Let us find its magnetic field at a point P at a perpendicular distance a from the point Q of the conductor, i.e. PQ = a <br> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/ARH_CHSE_ODI_13Y_SP_PHY_XII_C06_E01_064_S01.png" width="80%"> <br> Let a small current element d1 of the conductor at point O. Let the distance of d1 from Q be l, i.e. OQ = l. Let r be the position vector of point P relative to the current element d1 and `theta` be the angle between d1 and r. <br> From Biot-Savart.s law, the magnitude of field dB due to current element d1 will be <br> `dB=mu_(0)/(4pi)*(Idlsintheta)/r^(2)` <br> From right `DeltaOQP`. <br> `theta+phi=90^(@)rArrtheta=90^(@)-phi` <br> `thereforesintheta=sin(90^(@)-phi)=cosphi` <br> Also, `cosphi=a/r` <br> So, `r=a/cosphi=asecphi` <br> As, `tanphi=l/a` <br> `thereforel=atanphi` <br> On differentiating, we get <br> `dl=asec^(2)phidphi` <br> Hence, `dB=mu_(0)/(4pi)(I(asec^(2)phidphi)cosphi)/(a^(2)sec^(2)phi)` <br> `dB=(mu_(0)I)/(4pia)cosphidphi` <br> The direction of magnetic field due to all such current elements will be normally into the plane of paper. <br> The total field B due to the entire conductor will be <br> `B=int_(-phi_(1))^(phi_(2))dB=(mu_(0)I)/(4pia)int_(-phi_(1))^(phi_(2))cosphidphi` <br> = `(mu_(0)I)/(4pia)[sinphi]_(-phi_(1))^(phi_(2))=(mu_(0)I)/(4pia)[sinphi_(2)-sin(-phi_(1))]` <br> Hence, `B=(mu_(0)I)/(4pia)[sinphi_(1)+sinphi_(2)]`