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(D) The magnetic field $B$ at a distance $d$ from a finite straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{4\pi d}(\sin 	heta_1 +

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$1.5 	imes 10^{-5}\, T$

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(D) The magnetic field $B$ at a distance $d$ from a finite straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{4\pi d}(\sin 	heta_1 + \sin 	heta_2)$.Given,length of wire $AB = 6\, cm$,so the distance from the center to each end is $3\, cm$.The perpendicular distance $d$ from point $P$ to the wire is $4\, cm$ (from the $3-4-5$ triangle geometry).Here,$	heta_1 = 	heta_2 = 	heta$,where $\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{3}{5} = 0.6$.Thus,$B = \frac{\mu_0 I}{4\pi d}(2 \sin 	heta)$.Substituting the values: $B = \frac{(10^{-7} 	imes 4\pi) 	imes 5}{4\pi 	imes (4 	imes 10^{-2})} 	imes 2 	imes \frac{3}{5}$.$B = \frac{10^{-7} 	imes 5}{4 	imes 10^{-2}} 	imes \frac{6}{5} = \frac{10^{-5} 	imes 5}{4} 	imes 1.2 = 1.25 	imes 1.2 	imes 10^{-5} = 1.5 	imes 10^{-5}\, T$.

Find the magnetic field at point $P$ due to a straight line segment $AB$ of length $6\, cm$ carrying a current of $5\, A$. (See figure) $(\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A)$

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