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Question

A generator connected to the wheel or hub of a bicycle can be used to power lights or small electronic devices. A typical bicycle generator supplies 6.00 V when the wheels rotate at ω=20.0\omega=20.0 rad/s. If the generator's magnetic field has magnitude B = 0.600 T with N = 100 turns, find the loop area A.

Solution

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Answered 2 years ago
Answered 2 years ago
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The maximum emf induced in the coil is

εmax=NBAω\varepsilon_{max}=N B A \omega

We can express AA from above expression of maximum induced emf as

A=εNBω\begin{equation}A=\dfrac{\varepsilon}{NB\omega} \end{equation}

where, εmax=6 V\varepsilon_{max}=6\ \text{V} is the emf, B=0.600 TB=0.600\ \text{T} is the magnetic field, N=100 turnsN=100\ \text{turns} is the nnumber of turns, ω=20 rads\omega=20\ \frac{\text{rad}}{\text{s}} is the angular velocity.

Computing the above value in equation (1), we get

A=εmaxNBω=6 V(100 turns)(0.600 T)(20 rads)=0.005 m2\begin{align*} A & = \dfrac{\varepsilon_{max}}{NB\omega}\\ & = \dfrac{6\ \text{V}}{(100\ \text{turns})(0.600\ \text{T})(20\ \frac{\text{rad}}{\text{s}})}\\ & = 0.005\ \text{m}^2 \end{align*}

Therefore, the area of the loop is A=0.005 m2\boxed{A=0.005\ \text{m}^2}

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