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 $Bil=mg$ $B=\frac{mg}{il}=\frac{0.2\times 9.8}{2\times 1.5}=0.65T$