Magnetic field change for induced current

In summary: Yea, it's ambiguous. But it does seem likely that 10 is actually 10. Thanks, the round off error was very real.
  • #1
Sho Kano
372
3

Homework Statement


A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 10 cm formed from wire of diameter 2.5 mm and resistivity 8.5×10-8 Ω·m. At what rate must the magnitude of the magnetic field change to induce a 10 A current in the loop?

Homework Equations


[itex]\phi \quad =\quad BA,\quad B\bot A\\ { V }_{ induced }\quad =\quad \frac { d\phi }{ dt } \quad =\quad IR\\ R\quad =\quad \frac { \rho { L }_{ 2 } }{ { A }_{ 2 } } [/itex]

The Attempt at a Solution


[itex]Givens:\\ \\ { d }_{ 1 }\quad =\quad 10/100\quad =\quad 0.1\quad meters\\ { r }_{ 1 }\quad =\quad \frac { { d }_{ 1 } }{ 2 } \quad =\quad 0.05\quad meters\\ { d }_{ 2 }\quad =\quad 2.5/1000\quad =\quad 0.0025\quad meters\\ { r }_{ 2 }\quad =\quad \frac { { d }_{ 2 } }{ 2 } \quad =\quad 0.0013\quad meters\\ { L }_{ 2 }\quad =\quad 2\pi { r }_{ 2 }\quad =\quad 0.0082\quad meters\\ { A }_{ 1 }\quad =\quad \pi { { r }_{ 1 } }^{ 2 }\quad =\quad 0.0079\quad sqr\quad meters\\ { A }_{ 2 }\quad =\quad { \pi { { r }_{ 2 } } }^{ 2 }\quad =\quad 5.31\quad x\quad { 10 }^{ -6 }\quad sqr\quad meters\\ I\quad =\quad 10A\\ \rho \quad =\quad 8.5\quad x\quad { 10 }^{ -8 }\quad \Omega \quad meters[/itex]

[itex]Calculations:\\ \frac { d\phi }{ dt } \quad =\quad { A }_{ 1 }\frac { dB }{ dt } \quad =\quad IR\quad =\quad I\frac { \rho { L }_{ 2 } }{ { A }_{ 2 } } \\ { A }_{ 1 }\frac { dB }{ dt } \quad =\quad I\frac { \rho { L }_{ 2 } }{ { A }_{ 2 } } \\ \frac { dB }{ dt } \quad =\quad I\frac { \rho { L }_{ 2 } }{ { A }_{ 2 }{ A }_{ 1 } } \quad =\quad 0.1665\quad T/s[/itex]

Which is wrong?
 
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  • #2
Does your ##L_2## represent the correct length?
 
  • #3
TSny said:
Does your ##L_2## represent the correct length?
You are right, so L2 should be be done with r1? That gives me 0.3 meters
 
  • #4
Yes. Make sure you keep enough significant figures in the intermediate calculations so that you don't produce too much "round off error".
 
  • #5
TSny said:
Yes. Make sure you keep enough significant figures in the intermediate calculations so that you don't produce too much "round off error".
Gotcha, always keep sig figs. I'm now getting for the final answer 6.08 T/s Is that right?
 
  • #6
Sho Kano said:
I'm now getting for the final answer 6.08 T/s Is that right?
I'm getting closer to 7 T/s. I suspect you have too much round off error. Go back and recalculate ##r_1##, ##r_2##, ##A_1##, ##A_2##, and ##L## to at least 3 significant figures. Or, better, express your final answer symbolically in terms of the given quantities and then plug in the numbers only at the end of the calculation.
 
Last edited:
  • #7
TSny said:
I'm getting closer to 7 T/s.
Using the same numbers?
i = 10
p = 8.5 10-8
L = 0.3
A1 = 0.0079
A2 = 5.31 10-6
dB/dt = (i*p*L)/(A1*A2) = 6.0788 T/s
 
  • #8
I added an additional comment at the end of my last post. See if it helps.
 
  • #9
TSny said:
I added an additional comment at the end of my last post. See if it helps.
Okay I got it now,
[itex]\frac { dB }{ dt } \quad =\quad I\frac { \rho { L }_{ 1 } }{ { A }_{ 1 }{ A }_{ 2 } } \\ \frac { dB }{ dt } \quad =\quad I\frac { \rho 2\pi \frac { { d }_{ 1 } }{ 2 } }{ { \pi }^{ 2 }{ (\frac { { d }_{ 1 } }{ 2 } ) }^{ 2 }{ (\frac { { d }_{ 2 } }{ 2 } ) }^{ 2 } } \\ \frac { dB }{ dt } \quad =\quad I\frac { 16\rho \pi { d }_{ 1 } }{ { \pi }^{ 2 }{ { d }_{ 1 } }^{ 2 }{ d }_{ 2 }^{ 2 } } \quad =\quad 6.9264\quad =\quad 7\quad \frac { T }{ s } [/itex]

Thanks!
 
  • #10
OK. Since the numbers given in the problem have two significant figures, your answer should also be expressed to 2 significant figures: 6.9 T/s.
 
  • #11
TSny said:
OK. Since the numbers given in the problem have two significant figures, your answer should also be expressed to 2 significant figures: 6.9 T/s.
The diameter of 10 cm has 1 sig fig right, so shouldn't it have 1 sig fig?
 
  • #12
Well, 10 cm is ambiguous in terms of whether the 0 is counted as significant, unless your instructor has his or her own rules for that. Same for the 10 A. Since the other numbers are given to 2 sig figs, it suggests to me that the 10's also are to 2 significant figures. But, use your own judgment.
 
  • #13
TSny said:
Well, 10 cm is ambiguous in terms of whether the 0 is counted as significant, unless your instructor has his or her own rules for that. Same for the 10 A. Since the other numbers are given to 2 sig figs, it suggests to me that the 10's also are to 2 significant figures. But, use your own judgment.
Yea, it's ambiguous. But it does seem likely that 10 is actually 10. Thanks, the round off error was very real.
 

What is a magnetic field change?

A magnetic field change refers to any alteration in the strength or direction of a magnetic field. This can be caused by various factors, including the movement of electric charges or the presence of other magnetic fields.

What is an induced current?

An induced current is an electric current that is created in a conductor by a changing magnetic field. This phenomenon is known as electromagnetic induction and is the basis for many important technologies, such as generators and transformers.

How does a magnetic field change induce current?

When a magnetic field changes, it creates a force on charged particles in a conductor. This force causes the particles to move, creating an electric current. The direction of the induced current is always perpendicular to the direction of the changing magnetic field.

What are some practical applications of induced current?

Induced current has many practical applications, including generating electricity in power plants, powering motors and generators, and transmitting signals in devices like radios and televisions. It is also used in medical imaging technologies, such as MRI machines.

Can induced currents be harmful?

Yes, induced currents can be harmful under certain circumstances. For example, high levels of induced current can cause electric shocks, damage electronic devices, and interfere with sensitive equipment. Proper precautions must be taken when working with strong magnetic fields to avoid potential harm.

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