A wheel with 10 metallic spokes each 0.5 m long rotated with a speed of 120 rpm in a plane normal to the horizontal component of earth’s magnetic field Bh at a place. If Bh = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? (1G=10⁻⁴ T)choose the correct option from the given options.
(A) 0 V
(B) 0.628 mV
(C) 0.628 μV
(D) 62.8 μV
Answers
Dear Student,
◆ Answer -
(D) 62.8 μV
◆ Explanation -
Potential difference generated between rim and axis of cycle is calculated by -
EMF = 1/2 B.ω.l²
EMF = 1/2 × B × 2πf × l²
EMF = B × πf × l²
EMF = 0.4×10^-4 × 3.14 × 2 × 0.5^2
EMF = 62.8×10^-6 V
EMF = 62.8 μV
Therefore, EMF generated between rim and axis of cycle is 62.8 μV.
Thanks dear...
Explanation:
Length of metallic spoke = 0.5m (Given)
Speed = 120 rpm (Given)
Bh = 0.4
Guass = 4 x 10-5 T
Each spoke will act as a battery that is connected parallely between the two terminals. Thus, the emf produced by a rod moving with a velocity which is perpendicular to a magnetic field is - l.B x v
Therefore,
EMF = 1/2 × B × 2πf × l²
EMF = B × πf × l²
EMF = 0.4 × 10-4 × 22/7 × 2 × 0.5²
EMF = 6.3 × 10-5
Thus, the EMF generated will be 62.8 μV.