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NikAS [45]
3 years ago
7

Move the magnet at a relatively constant frequency back and forth through the coil. The voltage displayed is proportional to the

current flowing in the coil. What happens as you move the magnet through the coils with different number of loops?
Physics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

<em>The induced EMF, and hence the induced current produced will increase or decrease with the number of loops on the coil.</em>

Explanation:

According to Faraday' law of electromagnetic induction, the induced EMF increases with the speed with which we turn the coil, the surface area of the coil, the number of loops, and the strength of the magnetic field. From this, we can see that increasing the number of loops also increases the surface area involved. This means that if we move the magnet in this experiment through the coils with different number of loops, <em>the induced EMF, and hence the induced current, will increase or decrease with an increase or decrease  in the number of loops respectively. </em>

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A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi
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Answer:

5.15348 Beats/s

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Explanation:

v_1 = Velocity of sound = 342 m/s

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Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

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