A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if, (a) the wire intersects the axis, (b) the wire is turned from N-S to northeast-northwest direction, (c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Explanation

(a)
Here, length of wire inside the magnetic field is ab = 1 = 2r
`thereforel=(2)(0.1)=0.2m`
Magnetic force exerted on this wire is,
`F=IlBsintheta`
= `(7)(0.2)(1.5)sin90^(@)`
F = 2.1 N
(Vertically downward according to Ampere.s law `vecF=I(veclxxvecB)`)
(b)
Now in this case, length of conducting wire is `vec(cd)=vec(l.)` and it makes angle `theta` with magnetic field `vecB`.
Now in this case, length of conducting wire inside the cylindrical region of magnetic field is `vec(cd)=vec(l.)` which makes angle `theta` with the magnetic field.
Here magnetic force exerted on l. length of conducting wire is,
`F.=Il.Bsintheta`
= `IB(l.sintheta)`
= `IB(2r)` (From figure)
= (7)(1.5)(0.2)
`thereforeF.=2.1N`
( c)
As shown in the figure, wire is lowered by og = 6 cm. Now, the length of wire inside the cylindrical region of magnetic field is `fh=fg+gh=x+x=2x` which is perpendicular to `vecB`. Thus, `l_(_|_)=lsintheta=2x`.
In right angled `Deltaogf`,
`x=sqrt((10)^(2)-(6)^(2))=8cm`
`rArrlsintheta=2x=2xx8cm=16cm=0.16m`
`rArr` Magnetic force exerted on length fh will be,
`F..=IlBsintheta`
= `IB(lsintheta)`
= IB(2x)
= (7)(1.5)(0.16)
`thereforeF..=1.68N`