A long straight wire carrying a current of 30 | vedclass

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(C) Given: Current $i = 30 ~A$,External magnetic field $B_1 = 4 	imes 10^{-4} ~T$,Distance $r = 2.0 ~cm = 2 	imes 10^{-2} ~m$.The magnetic field pro

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$5 	imes 10^{-4}$

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(C) Given: Current $i = 30 ~A$,External magnetic field $B_1 = 4 	imes 10^{-4} ~T$,Distance $r = 2.0 ~cm = 2 	imes 10^{-2} ~m$.The magnetic field produced by the straight wire at a distance $r$ is given by $B_2 = \frac{\mu_0 i}{2 \pi r}$.Substituting the values: $B_2 = \frac{2 	imes 10^{-7} 	imes 30}{2 	imes 10^{-2}} = 3 	imes 10^{-4} ~T$.Since the external magnetic field $B_1$ is parallel to the current,it is perpendicular to the magnetic field $B_2$ produced by the wire (which circles the wire).The resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.$B = \sqrt{(4 	imes 10^{-4})^2 + (3 	imes 10^{-4})^2} = \sqrt{16 	imes 10^{-8} + 9 	imes 10^{-8}} = \sqrt{25 	imes 10^{-8}} = 5 	imes 10^{-4} ~T$.

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

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Match the following lists:
List-$I$ List-$II$
$(A)$ Benzene $1$. Phosgene
$(B)$ Ethylene $2$. Silver mirror
$(C)$ Acetaldehyde $3$. Mustard gas
$(D)$ Chloroform $4$. $(4n + 2) \pi$ electrons
$5$. Carbylamine

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If a point $(x, y) = (\tan \theta + \sin \theta, \tan \theta - \sin \theta)$,then the locus of $(x, y)$ is

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