A long wire carrying a current of 18 ,A is ke | vedclass

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(D) Given:Current in the long wire, $I = 18 \,A$.Magnetic field due to the solenoid, $B_1 = 8.0 	imes 10^{-3} \,T$ (directed along the axis).Distance

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$10 	imes 10^{-3} \,T$

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(D) Given:Current in the long wire, $I = 18 \,A$.Magnetic field due to the solenoid, $B_1 = 8.0 	imes 10^{-3} \,T$ (directed along the axis).Distance of point $P$ from the axis, $r = 0.6 \,mm = 0.6 	imes 10^{-3} \,m$.The magnetic field due to the long current-carrying wire at a distance $r$ is given by:$B_2 = \frac{\mu_0 I}{2 \pi r} = \frac{2 	imes 10^{-7} 	imes 18}{0.6 	imes 10^{-3}} = \frac{36 	imes 10^{-7}}{0.6 	imes 10^{-3}} = 60 	imes 10^{-4} \,T = 6 	imes 10^{-3} \,T$.The magnetic field $B_1$ due to the solenoid is along the axis, and the magnetic field $B_2$ due to the wire is tangential to the circle of radius $r$ around the wire. Thus, $B_1$ and $B_2$ are perpendicular to each other.The resultant magnetic field $B$ is:$B = \sqrt{B_1^2 + B_2^2} = \sqrt{(8 	imes 10^{-3})^2 + (6 	imes 10^{-3})^2} \,T$$B = \sqrt{64 	imes 10^{-6} + 36 	imes 10^{-6}} \,T = \sqrt{100 	imes 10^{-6}} \,T = 10 	imes 10^{-3} \,T$.

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