
A circular wheel of \[24\] spokes is rotated at \[300\] rpm in a uniform magnetic field of $2 \times {10^{ - 3}}$Tesla. Length of each spoke is \[25\]cm and magnetic field is along the axis of the wheel, then the value of induced e.m.f between rim and center of the wheel is?
Answer
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Hint: To solve this problem utilize the formula based on the magnetic field direction and the induced e.m.f concept. Then find the value of the angular velocity using the given frequency. After this, obtain the average velocity by finding the velocity of the wheel at the rim and at the center. Finally, substitute all these values in the emf and magnetic field equation. By using the given values and the formulas the value of the induced emf can be found out.
Formula used:
$\omega = 2\pi f$
\[emf\; = Bvl\]
Complete step by step answer:
The given values in this question are;
Number of spokes\[ = N = 24\]
The magnetic field B$ = 2 \times {10^{ - 3}}T$
The length of the spoke\[ = 25\]cm$ = 25 \times {10^{ - 2}}$ m $\because 1m = 100cm$
Frequency of rotation of the wheel$ = 300$rpm$ = \dfrac{{300}}{{60}} = 5$rps $\because 1\min = 60\sec $
We know that $1$ revolution$ = 2\pi $rad
The angular frequency is given by,$\omega = 2\pi f$
Substituting the value of the given frequency in the angular frequency equation we get$\omega = 10\pi $
At the rim, the velocity of the spoke is equal to $\omega $r
Whereas at the axis its velocity$ = 0$
Hence the average velocity of the wheel$ = \dfrac{{0 + \omega r}}{2} = \dfrac{1}{2}\omega r$
To find the emf induced between the center and rim of the wheel;
We know that \[emf\; = Bvl\]
$ = 2 \times {10^{ - 3}} \times \dfrac{1}{2}\omega r \times 25 \times {10^{ - 3}}$
$ = (2 \times {10^{ - 3}} \times \dfrac{1}{2} \times 10\pi \times 25 \times {10^{ - 2}}) \times 25 \times {10^{ - 2}}$
$ = 1.96 \times {10^{ - 3}}$
$ = 1.96$ mV
Note: While solving the problem care must be taken to convert the given length in centimeter to the meter using the equation$1m = 100cm$. In addition, convert the frequency given in rate per minute in the question to rate per second using the equation$1\min = 60\sec $. Also, note that not all values mentioned in the question need to be considered while solving the problem. For example in this question, the number of strokes has not been used in the solution.
Formula used:
$\omega = 2\pi f$
\[emf\; = Bvl\]
Complete step by step answer:
The given values in this question are;
Number of spokes\[ = N = 24\]
The magnetic field B$ = 2 \times {10^{ - 3}}T$
The length of the spoke\[ = 25\]cm$ = 25 \times {10^{ - 2}}$ m $\because 1m = 100cm$
Frequency of rotation of the wheel$ = 300$rpm$ = \dfrac{{300}}{{60}} = 5$rps $\because 1\min = 60\sec $
We know that $1$ revolution$ = 2\pi $rad
The angular frequency is given by,$\omega = 2\pi f$
Substituting the value of the given frequency in the angular frequency equation we get$\omega = 10\pi $
At the rim, the velocity of the spoke is equal to $\omega $r
Whereas at the axis its velocity$ = 0$
Hence the average velocity of the wheel$ = \dfrac{{0 + \omega r}}{2} = \dfrac{1}{2}\omega r$
To find the emf induced between the center and rim of the wheel;
We know that \[emf\; = Bvl\]
$ = 2 \times {10^{ - 3}} \times \dfrac{1}{2}\omega r \times 25 \times {10^{ - 3}}$
$ = (2 \times {10^{ - 3}} \times \dfrac{1}{2} \times 10\pi \times 25 \times {10^{ - 2}}) \times 25 \times {10^{ - 2}}$
$ = 1.96 \times {10^{ - 3}}$
$ = 1.96$ mV
Note: While solving the problem care must be taken to convert the given length in centimeter to the meter using the equation$1m = 100cm$. In addition, convert the frequency given in rate per minute in the question to rate per second using the equation$1\min = 60\sec $. Also, note that not all values mentioned in the question need to be considered while solving the problem. For example in this question, the number of strokes has not been used in the solution.
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