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  • (a) A metallic rod of length   is moved perpendicular to its length with velocity v in a magnetic field <img alt="\vec{B}" src="https://entrancecorner.cod

(a) A metallic rod of length \imath  is moved perpendicular to its length with velocity v in
a magnetic field \vec{B} acting perpendicular to the plane in which rod moves. Derive the expression for the induced emf.
(b) A wheel with 15 metallic spokes each 60 cm long, is rotated at 360 rev/min in a plane normal to the horizontal component of earth’s magnetic field. The angle of dip at that place is  60^{\circ}. If the emf induced between rim of the wheel and the axle is 400 mV, calculate the horizontal component of earth’s magnetic field at the place.
How will the induced emf change, if the number of spokes is increased?

 

 

 

 
 
 
 
 

Answers (1)

Let the rod is moving with a constant velocity v in a magnetic field B perpendicular to the plane in which the rod moves.
Then the force experienced by the rod
F= ilB
if the rod moves a distance x then the work is done
dw= il B x and the potential
V=\frac{w}{q}= \frac{ql Bx}{t\times q}= Bl\frac{x}{t}= Bl V
b)
360\; \frac{rev}{min}= 6\; \frac{rev}{sec}= n
emf induced between rim of the wheel and axle
E= 400\times 10^{-3}V
Angle of dip= 60^{\circ}
induced

emf= \frac{1}{2}Bl ^{2}w
w= 2\pi n
emf\: E= \frac{1}{2}\times\, B\times \left ( 0\cdot 6 \right )^{2}\times 2\times \pi \times 6
B= \frac{2\times 400\times 10^{-3}}{0\cdot 36\times 2\times 6\times \pi }= 58\cdot 9\times 10^{-3}T
                                                = 0\cdot 0589T
B_{H}= B\cos \delta
       = 0\cdot 0589\cos 60
       = 0\cdot 02945\, T   

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Safeer PP

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