6.17 A line charge l 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?

6.17 A line charge l 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?

6.17 Line charge per unit length = = =
Where r = distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field, =
At a distance r, the magnetic force is balanced by the centripetal force. i.e.
BQ = , where v = linear velocity of the wheel. Then,
B =
v =
Angular
Similar Questions for you
Kindly go through the solution
Bv = B sin 60°
->
M = φ? /I? = (B? A? )/I? = [ (μ? I? /2R? )πR? ²]/I?
[Diagram of two concentric coils]
M = (μ? πR? ²)/ (2R? )
M ∝ R? ²/R?
(A) The magnet's entry
R =
L = 2 mH
E = 9V
Just after the switch ‘S’ is closed, the inductor acts as open circuit.
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Physics Ncert Solutions Class 12th 2023
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