A wheel with 12 metal spokes each 0.6 m long is rotated with a speed of 180 r.p.m in a plane normal to earth's magnetic field at a place. If the magnitude of the field is 0.6 G, what is the magnitude of induced e.m.f. between the axle and rim of the wheel? (Answer: 2.035 ｘ 10⁻⁴ V) 



Data Given:

Numbers of metal Spoke = N = 12 spokes

Length of each spoke = r = 0.6 m

Angular Velocity = ധ = 180 rpm = 180 ｘ `\frac {2ㄫ}{60}` rad s⁻¹ = 6ㄫ rad s⁻¹

Magnetic Field = B = o.6 G = 0.6 ｘ 10⁻⁴ T

To Find:

Induced e.m.f. = Ɛ = ?


Solution:

The formula for motional e.m.f is given by:

Ɛ = B l v  ----------------(1)

Where Length v and velocity v is unknown. let it be find first:

To Find Length l:


Circumference of the wheel will be = 2 ㄫ r = 2 ｘ 3.1416 ｘ 0.6 m = 3.7699 m 

Length between two spoke = l = circumference/numbers of spoke = `\frac {3.7699 m }{12}` = 0.314 m ------(2)


To find Velocity v we have

v = r ധ

by putting values

v = 0.6 m  ｘ 6ㄫ rad s⁻¹

v = 11.310 m s⁻¹ -------------(3)


Now putting the values of equations (2), (3), and value of B in equation (1) 

Ɛ = B l v

Ɛ = 0.6 ｘ 10⁻⁴ T ｘ 0.314 m ｘ 11.310 m s⁻¹

Ɛ = 2.131 ｘ 10⁻⁴  -----------------Ans.



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