A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. (a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero? (b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) `g = 9.8 m s^(-2)` .

Explanation

(a)
Suppose, magnetic field `vecB` is applied perpendicularly inside the plane of figure and also perpendicular to the length of the rod such that magnetic force exerted on the rod can balance its weight. Hence,
`IlB=mg`
`thereforeB=(mg)/(Il)`
`thereforeB=((0.06)(9.8))/((5)(0.45))=0.2613T`
In this situation, as shown in FBD figure (2), for vertical equilibrium of the rod,
`T.+IlB=mg`
`thereforeT.=0" "(because"Here",IlB=mg)`
(b)
Now, when above magnetic field is reversed if new total tension in the two suspension wires is T.. then for vertical equilibrium,
`T..=IlB+mg`
= `2mg" "(becauseIlB=mg)`
= (2) (0.06) (9.8)
`thereforeT..=1.176N`