A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the (a) total torque on the coil, (b) total force on the coil, (c) average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross-sectional area `10^(-5) m^2`, and the free electron density in copper is given to be about `10^29 m^(-3)` .)

Explanation

(a)
We have, `tau=BINAsintheta`
Here, `theta` = angle between `vecAandvecB=0^(@)`
`thereforetau=BINAsin0^(@)`
`thereforetau=0`
(b) Magnetic force on a current element `Ivec(dl)` is `vec(dF)=Ivec(dl)xxvecB` (According to Ampere.s law)
`thereforeintvec(dF)=Iintvec(dl)xxvecB`
`thereforevecF=I(vec0)xxvecB`
(`because` For a closed loop, `intvec(dl)=vec0`)
`thereforevecF=vec0xxvecB=vec0`
( c) Free electron no. density,
`n=N/V=N/(Adl)`
`therefore` Total no. of free electrons in volume V = Adl of length element wil be,
N = Andl
Magnetic force exerted on above number is,
`dF=IdlBsin90^(@)" "("From diagram "vec(dl)_|_vecB)`
`thereforedF=IdlB`
Magnetic force on one free electron will be,
`dF.=(dF)/N=(IdlB)/(Andl)=(IB)/(An)`
`thereforedF.=((5)(0.1))/((10^(-5))(10^(29)))`
`thereforedF.=5xx10^(-25)N`