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December 9th, 2018, 04:33 AM   #1
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wire frames as a dynamo

Hi

I have the query of the following nature:

In Fig. 1, a wire frame ABCD) is drawn in a homogeneous magnetic field B = 0.10 T . The points P and Q are connected to a voltmeter. Then the wire frame is mechanically driven. In this case side AD moves forward in the position shown (paper out). The wire frame has a surface of 60 cm2 and functions as a dynamo.

Query 1: Find out in which direction a current flows through the wire frame during the first quarter turn of the wire frame.

Figure 2
In figure 2 a graph of the flux 'phi' through the wire frame is drawn in the course of time. The flux can also be calculated by drawing up a sinus formula.

Query 2 : Determine the maximum voltage of this 'dynamo'.

The rotational frequency of the dynamo is reduced to half.

Query 2. Sketch the corresponding (* U ind, t) graph. **where U is Indution and is subscript ind. Thus U subscript 'ind'.
Give the characteristic points in the graph clearly with corresponding values.

++++++++++++++++++++++++++++++++++++++++++++++++++ ++++

can anyone give me some pointers on how to tackle this situation.
some feedback on how to proceed would be of a good assistance here.

Thanks in advance
Attached Images Diagram 1.jpg (9.9 KB, 2 views) Diagram 2.jpg (16.4 KB, 24 views) December 9th, 2018, 10:36 AM   #2
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Quote:
 Originally Posted by hello_math In Fig. 1, a wire frame ABCD) is drawn in a homogeneous magnetic field B = 0.10 T . The points P and Q are connected to a voltmeter. Then the wire frame is mechanically driven. In this case side AD moves forward in the position shown (paper out). The wire frame has a surface of 60 cm2 and functions as a dynamo. Query 1: Find out in which direction a current flows through the wire frame during the first quarter turn of the wire frame.
Use the left hand rule for generators (see attached thumbnail diagram). Note the induced current is maximum when the wire frame is at the position shown (moving forward), and zero when the wire frame is 90 degrees (quarter turn) from the initial position.

Quote:
 Figure 2 In figure 2 a graph of the flux 'phi' through the wire frame is drawn in the course of time. The flux can also be calculated by drawing up a sinus formula. Query 2 : Determine the maximum voltage of this 'dynamo'. from the sinusoid, note the period $T = 0.04 \, s$ and the amplitude is $\Phi = 0.0006 \, Wb$ ...

$\Phi (t) = 0.0006\sin(50\pi \cdot t)$

$\dfrac{d \Phi}{dt} = 0.03\pi \cos(50\pi \cdot t)$

Max voltage occurs when $\dfrac{d \Phi}{dt}$ is a maximum $\implies \cos(50\pi \cdot t) = 1$

also note (for a single looped wire frame) $E = -\dfrac{d \Phi}{dt}$

Quote:
 The rotational frequency of the dynamo is reduced to half.
The graph of flux at a different frequency will have its period changed ... note period, $T$, is inversely proportional to frequency, $f$.

How will changing the frequency to half its initial value affect the period of the graph?
Attached Images Left_rule.jpg (65.7 KB, 2 views) December 9th, 2018, 11:02 AM #3 Member   Joined: Nov 2012 From: The Netherlands Posts: 40 Thanks: 2 Thank you skeeter This will at least help me to kickstart the issue. Let me dwell into this and I fill post my side of the feedback soon. Thanks ! December 9th, 2018, 12:29 PM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 wb Skeeter, good to see you again. Thanks from greg1313 and skeeter December 20th, 2018, 01:49 AM #5 Member   Joined: Nov 2012 From: The Netherlands Posts: 40 Thanks: 2 Hi All as I had stated, I would come back and give the final feedback. (It's not me ) but my son has been able to do what was required (I think). The resolution is as here : https://i.postimg.cc/NFXNJ2pN/problms.jpg This was just for the records. Thanks Tags dynamo, frames, wire Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post EBTERTTBT Computer Science 12 September 4th, 2018 11:41 PM symmetry Algebra 1 June 12th, 2007 09:12 PM

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