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JEE Main 2024 (30 Jan, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2024 (30 Jan, Shift 1)

Q16JEE Main 2024 · 30 JAN · SHIFT 1Medium

Force on Current-Carrying Conductors

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

Numerical answer type. Work it out, then open the answer.

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Answer: 35

  1. BHB_H points towards geographic north; the wire points from south-east to north-west.
  2. Angle between north-west and north is 45∘45^\circ.
  3. Fl=IBHsin⁡45∘=2×3.5×10−5×12\frac{F}{l} = I B_H \sin45^\circ = \sqrt{2}\times3.5\times10^{-5}\times\frac{1}{\sqrt{2}}
  4. Fl=3.5×10−5=35×10−6 N/m\frac{F}{l} = 3.5\times10^{-5} = 35\times10^{-6}\text{ N/m}, so the answer is 35.
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