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Question: A bicycle wheel of radius \[0.5{\rm{ }}m\]has \[32\] spokes. It is rotating at the rate of \[120\]re...

A bicycle wheel of radius 0.5m0.5{\rm{ }}mhas 3232 spokes. It is rotating at the rate of 120120revolutions per minute, horizontal component of earth of earth’s magnetic field βH=4×10−5T{\beta _H} = 4 \times {10^{ - 5}}T. The emf induced between the rim and the centre of wheel will be:

A.6.28×10−5V6.28 \times {10^{ - 5}}V
B.4.8×10−5V4.8 \times {10^{ - 5}}V
C.6×10−5V6 \times {10^{ - 5}}V
D.1.6×10−5V1.6 \times {10^{ - 5}}V

Explanation

Solution

An electromotive force, is the electrical action produced by a non-electrical source. Hence emf is directly proportional to magnetic field, area swept by spoke and revolutions made by wheel per second.

Complete Step by step answer: Radius = r=0.5mr{\rm{ }} = {\rm{ }}0.5{\rm{ }}m
Earth’s magnetic field βH=4×10−5T{\beta _H} = 4 \times {10^{ - 5}}T.
Number of spokes = n=32n{\rm{ }} = {\rm{ }}32
Number of rotations completed per minutes by wheel =120 = {\rm{ }}120
Hence rotations completed per second by a wheel = 12060=2\dfrac{{120}}{{60}} = 2 revolutions per second.
Since radius of the wheel is 0.5m0.5m, the length of each spoke = 0.5m0.5m
The rate of change of an angle with respect to velocity is called angular velocity. The SI unit of angular velocity id radian per second.
Hence angular velocity = w=2πnw = 2\pi n
In time period of TTthe area swept by one spoke = πr2\pi {r^2}
The time period of TTthe area swept by one spoke = π×(0.5)2\pi \times {(0.5)^2}= 0.25πm20.25\pi {m^2}
The emf induced in the e=dϕdte = \dfrac{{d\phi }}{{dt}}
The emf induced in the e=d(BA)dte = \dfrac{{d(BA)}}{{dt}}………………………………………………………….. (I)
Where BB= magnetic field and
AA= surface area
We can write equation (I) as e=BATe = \dfrac{{BA}}{T}………………………………………………………. (II)
Since the frequency is the number of occurrences of a repeating event per unit of time,
T=1fT = \dfrac{1}{f}
⇒f=w2π\Rightarrow f = \dfrac{w}{{2\pi }}
⇒f=4π2π\Rightarrow f = \dfrac{{4\pi }}{{2\pi }}
⇒f=2Hz\Rightarrow f = 2Hz
Now, we can write equation (II) as:
e=BAfe{\rm{ }} = {\rm{ }}BAf
Means emf is directly proportional to magnetic field, area swept by spoke and revolutions made by wheel per second.
⇒e=4×10−5×0.25π×2\Rightarrow e = 4 \times {10^{ - 5}} \times 0.25\pi \times 2
⇒e=6.28×10−5V\Rightarrow e = 6.28 \times {10^{ - 5}}V

Therefore the emf induced between the rim and the centre of the wheel will be = e=6.28×10−5Ve = 6.28 \times {10^{ - 5}}V. So, the correct option is A.

Note: E.M.F always opposes the change in magnetic flux associated with a conducting loop. Depending on the direction of the induced current it can be signed negative or positive.