Question

A 2.5 m long straight wire having mass of 500 g is suspended in mid air by a uniform horizontal magnetic field B. If a current of 4 A is passing through the wire then the magnitude of the field is then (Take g = 10 $m{s}^{-2}$)

• A. 0.5 T
• B. 0.6 T
• C. 0.25 T
• D. 0.8 T

Answer 1

The correct option is A 0.5 T

Here, $m=500g=0.5kg,I=4A,l=2.5m$
As $F=IlBsin\theta$

As the rod is suspended in air so,weight of thr rod is balanced by the magnetic force acting on the current carrying rod
$mg=IlBsin{90}^o$, $(\because \theta ={90}^{o}andF=mg)$
$\therefore B=\frac{mg}{Il}=\frac{0.5\times 10}{4\times 2.5}=0.5T$