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A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. To suspend it in air by a uniform horizontal magnetic field, the value of the required magnetic field is ______ T.

Asked in GSEB Board March 2020 · Force on a wire in a field

• (1) 0.45
• (2) 6.5
• (3) 0.65
• (4) 4.5

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives 0.45 T. Sanity check: the balance BIℓ=mg fixes B=(0.2×9.8)/(2×1.5)=0.65 T.
• (2) 6.5 T is ten times too large: it is what the mass taken as 2 kg gives (reading the length as 0.15 m does the same). 200 g is 0.2 kg.
• (4) 4.5 T is ten times an already unreachable value. With m=0.2 kg, ℓ=1.5 m and I=2 A the balance leaves no freedom: B=0.65 T.

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