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(C) The magnetic field $B$ due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1 + \si

Vedclass · Accepted Answer

$0.4$

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(C) The magnetic field $B$ due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.Given: $L = 32 \,cm$, so half-length $a = 16 \,cm$. Perpendicular distance $r = 12 \,cm$.The hypotenuse $d = \sqrt{r^2 + a^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \,cm$.Thus, $\sin 	heta_1 = \sin 	heta_2 = \frac{a}{d} = \frac{16}{20} = 0.8$.Substituting the values:$B = \frac{10^{-7} 	imes 30}{12 	imes 10^{-2}} (0.8 + 0.8)$$B = \frac{30 	imes 10^{-5}}{12} (1.6) = 2.5 	imes 10^{-5} 	imes 1.6 = 4 	imes 10^{-5} \,T$.Since $1 \,T = 10^4 \,G$, we have $B = 4 	imes 10^{-5} 	imes 10^4 \,G = 0.4 \,G$.

$A$ straight conductor of length $32 \,cm$ carries a current of $30 \,A$. Magnetic induction at a point in air at a perpendicular distance of $12 \,cm$ from the mid-point of the conductor is (in $G$)

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