**Question:**

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. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B = − B0 k (*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 $=\lambda=\frac{\text { Total charge }}{\text { Length }}=\frac{Q}{2 \pi r}$

Where,

*r *= Distance of the point within the wheel

Mass of the wheel = *M*

Radius of the wheel = *R*

Magnetic field, $\vec{B}=-B_{0} \hat{k}$

At distance *r*,themagnetic force is balanced by the centripetal force i.e.,

$B Q v=\frac{M v^{2}}{r}$

Where,

$v=$ linear velocity of the wheel

$\therefore B 2 \pi r \lambda=\frac{M v}{r}$

$v=\frac{B 2 \pi \lambda r^{2}}{M}$

$\therefore$ Angular velocity, $\omega=\frac{v}{R}=\frac{B 2 \pi \lambda r^{2}}{M R}$

For $r \leq$ and $a

$\omega=-\frac{2 \boldsymbol{A}_{0} \quad a^{2} \lambda}{M R} \hat{k}$