Question

A straight wire of mass 200 g and length $ \text{1}.\text{5 m} $ carries a current of 2 A. It is suspended in mid air by a uniform horizontal magnetic field B, The magnitude of B (in tesla) is

• A. $ 0.\text{65} $
• B. $ 0.\text{55} $
• C. $ 0.\text{75} $
• D. $ 0.\text{45} $

Answer 1

Answer: (A)

Given, mass of the wire = 200 g = 02 kg Length of the wire = 15 m Current $ i=2A $ Magnitic field B =? The force acting on the current carrying wire in uniform magnetic field $ F=Bil\sin \theta $ $ F=Bil $ $ (\because \theta ={{90}^{\circ }}) $ Weight of the wire w = mg = 0.2 $ \times $ 9.8N In the position of suspension $ Bil=mg $ $ B=\frac{mg}{il}=\frac{0.2\times 9.8}{2\times 1.5}=0.65T $