A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R.
Question:

A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass 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,

= − B0 (≤ aR)

= 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,

= 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 $a<R$, we get:

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

 

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