A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid air by a uniform horizontal magnetic field B (figure). What is the magnitude of the magnetic field ?

Explanation

1.
Consider a uniform magnetic field `vecB`, applied perpendicularly inside the plane of figure. A conducting wire of length l is kept perpendicular to `vecB`. When a current I is passed through this wire from left to right, it experiences a magnetic force,
`vecF_(m)=I(vecIxxvecB)`
`thereforeF_(m)(uarr)=IlBsin90^(@)" "(becausevecl_|_vecB)`
`thereforeF_(m)(uarr)=IlB`
2. For mid-air suspension, above force must be getting balanced by gravitational force acting on wire in vertically downward direction. Hence,
`F_(m)(uarr)=F_(g)(darr)`
`thereforeIlB=mg`
`thereforeB=(mg)/(Il)`
`thereforeB=((0.2)(9.8))/((2)(1.5))`
`thereforeB=0.6533` T