![A line charge A 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 = – Bo k (r ≤ a; a < R) = 0 (otherwise].What is the angular velocity of the wheel after the field is suddenly switched off?](https://discussion.tiwariacademy.com/app/uploads/2021/03/Fig-6.22.png)
Class 12 Physics
CBSE and UP Board
Electromagnetic Induction
Chapter-6 Exercise 6.17
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 6 Question 17
A line charge A 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 = – Bo k (r ≤ a; a < R) = 0 (otherwise].What is the angular velocity of the wheel after the field is suddenly switched off?
Share
Line charge per unit length = λ = Total Charge/Length = Q/2πr
Where, r = Distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field, B = -B0 k
At distance r, the magnetic force is balanced by the centripetal force i.e.,
BQv = Mv²/r
Where,
v = linear velocity of the wheel
Therefore, B2πrλ = Mv/r
v =B2πλr²/M
Therefore ,Angular velocity , ω =v/R = B2πλr²/MR
For r ≤ a ≤ R, we get
ω = -(2πB0a²λ/MR) k^