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The horizontal component of the Earth's magnetic field at a certain plate is 3×10⁻⁵ T and the direction of the field is from the geographic South to the geographic North. A very long straight conductor is carrying a steady current of 1 A. What is the force per unit length on it when it is placed on a horizontal table and the direction of current is South to North?

Asked in GUJCET 2025 · Field of a bar magnet and earth's field

• (1) Zero
• (2) 1×10⁻⁵ Nm⁻¹
• (3) 6×10⁻⁵ Nm⁻¹
• (4) 9×10⁻⁵ Nm⁻¹

Step-by-step solution

Why the other options are wrong

• (2) F/ℓ=BI sin θ vanishes here because θ=0, whatever the numbers are. And 1×10⁻⁵ is not even the largest possible value: that is BI=3×10⁻⁵ N m⁻¹, for a wire laid east to west.
• (3) 6×10⁻⁵ is 2BI - twice the largest force per unit length the wire could feel in this field, which is BI=3×10⁻⁵ N m⁻¹ and needs the wire perpendicular to the field. Here it is parallel, so the force is zero.
• (4) 9×10⁻⁵ is 3BI. No orientation of a 1 A wire in a 3×10⁻⁵ T field gives more than 3×10⁻⁵ N m⁻¹, and with the current along the field it gives nothing at all.

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