Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

197 of 276

Question 197 of 276

The horizontal component of earth's magnetic field at a place is 3.5×10−5T3.5\times {10}^{-5}T. A very long straight conductor carrying current of 2A\sqrt{2}A in the direction from South east to North West is placed. The force per unit length experienced by the conductor is ________×10−6Nm−1\times {10}^{-6}N{m}^{-1}.

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Answer: 35

Horizontal component of magnetic field, BH=3.5×10−5T{B}_{H}=3.5\times {10}^{-5}T
Force acting on wire, F=ilBsin⁡θ,wherei=2AF=ilB\sin \theta ,\text{where}i=\sqrt{2}A
Therefore, Fl=iBsin⁡θ=2×3.5×10−5×12\dfrac{F}{l}=iB\sin \theta =\sqrt{2}\times 3.5\times {10}^{-5}\times \dfrac{1}{\sqrt{2}}
=35×10−6Nm−1=35\times {10}^{-6}N{m}^{-1}

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