A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig a). A uniform magnetic field extends over a circular region within the rim. It is given by,
B = – B₀k (r ≤ a; a < R)
= 0 (otherwise)
What is the angular velocity of the wheel after the field is suddenly switched off?

Solution

Line charge per unit length is given by,

$λ = \frac{Totalcharge}{Length} = \frac{Q}{2 πr}$

where,
‘r’ is the distance of the point from the wheel,
M is the mass of the wheel,
R is the radius of the wheel, and
$\overset{⇀}{B} = - B_{o} \hat{k}$ is the magnetic field.

At distance r, the magnetic force is balanced by the centripetal force.

i.e. $QvB = \frac{{Mv}^{2}}{r}$

where, v is the linear velocity of the wheel.

$∴ B 2 πrλ = \frac{Mv}{r}$

$i . e ., ν = \frac{B 2 {πλr}^{2}}{M}$

Therefore,
Angular velocity, $ω = \frac{v}{r} = \frac{B 2 {πλr}^{2}}{MR}$

For $r ⩽ a; a < R,$ we get

$ω = - \frac{2 B_{o} a^{2} λ}{MR} \hat{k}$ ,

is the required angular velocity of the wheel after the field is suddenly switched off.

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