The horizontal component of Earth's magnetic | vedclass

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(A) The horizontal component of the Earth's magnetic field is $B_H = 3.5 	imes 10^{-5} \,T$.The current in the conductor is $i = \sqrt{2} \,A$.The co

Vedclass · Accepted Answer

$35$

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(A) The horizontal component of the Earth's magnetic field is $B_H = 3.5 	imes 10^{-5} \,T$.The current in the conductor is $i = \sqrt{2} \,A$.The conductor is placed from South-East to North-West,which makes an angle of $	heta = 45^\circ$ with the horizontal magnetic field (North-South direction).The force per unit length on a current-carrying conductor is given by $\frac{F}{\ell} = i B_H \sin 	heta$.Substituting the values: $\frac{F}{\ell} = \sqrt{2} 	imes (3.5 	imes 10^{-5}) 	imes \sin(45^\circ)$.Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$,we get $\frac{F}{\ell} = \sqrt{2} 	imes 3.5 	imes 10^{-5} 	imes \frac{1}{\sqrt{2}} = 3.5 	imes 10^{-5} \,N/m$.Converting to the required format: $3.5 	imes 10^{-5} = 35 	imes 10^{-6} \,N/m$.Therefore,the value is $35$.

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