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1. If the current in the wire inside a lightbulb is 3A, what is the current in the connecting wire going into the lightbulb? The connecting wire must have a current...

Question

1. If the current in the wire inside a lightbulb is 3A, what is the current in the connecting wire going into the lightbulb? The connecting wire must have a current of more than 3A The connecting wire must have a current of 3A The connecting wire must have a current of less than 3AWhich of the following best describes the source of all magnetic fields? Magnetic fields are due to magnetically charged particles Magnetic fields are due to the presence of metal atoms Magnetic fields are due to movin

1. If the current in the wire inside a lightbulb is 3A, what is the current in the connecting wire going into the lightbulb? The connecting wire must have a current of more than 3A The connecting wire must have a current of 3A The connecting wire must have a current of less than 3A Which of the following best describes the source of all magnetic fields? Magnetic fields are due to magnetically charged particles Magnetic fields are due to the presence of metal atoms Magnetic fields are due to moving charged particles Magnetic fields are due to A magnetic field can exert a force on an object; but only The object has an electric charge The object is in motion The object's motion is perpendicular to the direction of the magnetic field All of the above musts be true 4. If a current is running through a wire, which of the following will be induced around the wire? A reverse current gravitational field magnetic field A sound wave 5. If a proton is moving at a speed of 5 mls to the right (7 ) through a uniform magnetic field that points to the left ( <-) , which of the following describes the force the proton feels? The proton feels a force pushing it upward The proton feels a force pushing it downward The proton feels a force pushing it to the right The proton feels a force pushing it to the left The proton feels zero force



Answers

What objects experience a force in an electric field? Chapter 23 gives the answer: any electric charge, stationary or moving, other than the charge that created the field. What creates an electric field? Any electric charge, stationary or moving, as you studied in Chapter $23 .$ What objects experience a force in a magnetic field? An electric current or a moving electric charge, other than the current or charge that created the field, as discussed in Chapter $29 .$ What creates a magnetic field? An electric current, as you studied in Section $30.1,$ or a moving electric charge, as shown in this problem. (a) To display how a moving charge creates a magnetic field, consider a charge $q$ moving with velocity $\mathbf{v} .$ Define the vector $\mathbf{r}=r \hat{\mathbf{r}}$ to lead from the charge to some location. Show that the magnetic field at that location is
$$\mathbf{B}=\frac{\mu_{0}}{4 \pi} \frac{q \mathbf{v} \times \hat{\mathbf{r}}}{r^{2}}$$
(b) Find the magnitude of the magnetic field 1.00 $\mathrm{mm}$ to the side of a proton moving at $2.00 \times 10^{7} \mathrm{m} / \mathrm{s}$ . (c) Find the magnetic force on a second proton at this point, moving with the same speed in the opposite direction. (d) Find the electric force on the second proton.

For this problem On the topic of magnetic fields, we are told to consider a charge Q. Moving with velocity V. We want to find the magnetic field at some location which is a distance are away from the moving charge. And we are, we want to show that this memory field is more not over four pi times Q times V. Across our hat. Over R squared. We then want to find the magnitude of the magnetic field a millimeter to the side of a proton that is moving at two times 10 to the seven m per second. We don't want to find the magnetic force on a second. proton at this point moving with the same speed in the opposite direction. And then we want to find the electric force on the second proton. Now we know in the equation D. B is equal to mu notes over four pi r squared times the current I time's D s across our hat. The moving charge constitutes a bit of current, which is I. Is equal to end the Q. V. A. Where n is the charge carrier density cure? Is the total charge, the the velocity and a the area. Now for a positive charge. The direction the S. Is in the same direction of V. So D B. Is equal to um you know it over four pi R squared times in Q. A. Times the magnitude of Ds times the velocity vector we crossed with our hat. Now A. D. S. Is the volume occupied by the moving charge and N A D. S. Is equal to one for just one charge. And so therefore the total magnetic field B is equal to mu notes over four pi R squared times Q times V across our hat, which is the magnetic field that we require next For part B. We want to find the magnitude of the magnetic field one millimeter away from a proton that is moving at two times seven, two times 10 to the power seven m per second. And so using this equation we get B is equal to um you know it, which is four pi times 10 to the minus seven Tesla meter per ampere multiplied by the charge of the proton, 1.6 times 10 to the minus 19 columns, times the velocity, two times 10 to the seven meters the second. And this is multiplied by the sign of 90 degrees since D. S. And V. A. In the same direction or perpendicular, rather divided by full pie into one times 10 to the minus three, all squared. And so calculating, we get the magnetic field strength to be three 0.2 times 10 to the minus 13 kessler's for part C. We want to find the manually force on a second proton at this point that is moving with the same speed but in the opposite direction. Now, the magnetic force F. B is given by Q times the magnitude of the Crosby. And this is equal to the force on the proton, 1.6 times 10 to the minus 19 columns, times the velocity to times 10 to the seven m per second, times the magnetic field strength 3.2 times 10 to the minus 13 kessler's times the sine of 90 degrees. And so we get the magnetic force to be one 0.2 times 10 to the minus 24 newton's. And this is a way from the first proton. And lastly, for part D of this problem, we want to find the electric force on the second proton. Now the electric force is F. E. And this is given by Q times E, which is K. E. The electric constant times the two charges, Q one times Q two over R squared. This is the electric constant 8.99 times 10 to the nine newton meter squared McCullum squared times the two charges, which are both the charge on a proton, 1.6 times 10 to the minus 19 columns squared, divided by their separation won times 10 to the minus three m one millimeter squared. So we get the electric force to be two 0.3 times 10 to the minus 22 newtons. And again, this is away from the first proton.

To see something put. Mission is given in this problem. Your source off Electric Village Judge I did it. Rest own in motion. It doesn't matter. But sort off my great village moving judge are cutting. Not in this question. In the first part, we have toe calculate magnetic field intensity due to a movie judge and chicken parts. We have a local public value up magnetic will intensity do toe movie bit validity off doing 10 to the power seven. Meet up our ticket after white but millimeter promised spot in turn or C park Be able to calculate the ports off interaction between true prevalence, Moving it upload your direction and in the deeper they have to calculate the forethought. That is, if this is magnetic force electric ports, we have to calculate on the second proton due to past due to second on the second platoon that these are the the best part of this question which I put when we start solving with obstinate support charge you it moving along this but bit by little TV on. We are finding the magnetic field intensity at three point having w Shin bet that hot as you know that better. The very current element and I couldn't is flowing through it than magnetic field intensity at any point can be given by I be it because our unit vector divided by r squared Why should I loved this concept? Give will apply here. Let us see you the product off current and it's element be it's can be originate as you know Currently defined it Charge flowing body written right and be a support duty The velocity covered by the charge But in the time so here be your support beauty candidate in it velocity of the judge Here I am writing separately Felicity can be defined a the spits covered by the church But he did time So this concept we are using substitute this value Any question But the magnetic reading density at any point due to moving judge, we can write you not upon poor fight I can get in it You Are you in ideas in Britain as you we crossed be 30. Would you submit that are divided bite? How do you square? So this is about as there are a bust. So if any charges moving, then it will creates a magnetic field around and which could be measured by this bar. So this is very important relations. No integral part in group would want odd moving Cuddy in second part. If Proton is moving, been desperate up. Can do the power. 6 30 With that speed up, two into 10 to the power of it. Meet up our ticket. Then we have to calculate the magnetic. Me? Let me point. Which is that a separation off? Which is that the separation off? One millimeter That is 10 to the power off minus three. Bitter. We used the same relation. Magnetic field can be given. What you not. Upon bullfight, you re cross me, uh, unit acted upon. Are you square in our substitute to calculate the magnetic Quit. We're calculating here. Only the magnitude. So Ford by in tow tend to de power off minus seven upon port by charge on the protons. But 0.6 in 10 departments 90. The spirit is given two into 10 to the power seven on. Dispenses tend to depart my industry only square on bill. Between these two, it's 19. But this is a sign off. 90 on solving you will get Magarri feeling pencil. It will be 3.2 into Tend to deposit off 30 Islam. So this is the answer up. Second part? No. In the third particles given, unforgettably moving in this direction. Big publicity. Two into 10 to develop. Save it, meet a per second. And that is moving in a positive direction with that same validity at the separation off one millimeter. So you have to cooperate the force off interaction with me them before disability. You Sorry. Just a moment, please. Who is going to find it? She window the cross Me, They will find the magnitude charge on the protons spirits doing 10 to the power seven on magnetic will be every calculated on angle will be 90 degree. So this course you will get but white You don't go into 10 to the power up minus country for nuclear and they will depend each other. So they're moving in a positive direction. So they are repelling. So there will be weapons imports between that in their deeper we have to calculate forced to do electric field would be there. There will be electricity, so they will be there will be a port with them. Okay, Cuban butto by artists quit nine in 10 to the other night, 1.6 into 10 to the power off minus 90 1.6 in 10 to the power off 90. His but millimeters that it's tend to the poor by the three Holy square. So the port on. But we're done on other booby 2.3 and 10 to the power minus T. And they are repelling that This is the repelling puts here. I have to make a correction. I have written drunk because they are moving in a positive direction. The force will be attractive. I'm really very sorry. I am doing the correction here. Magnetic ports will be attractive, but electric port will be repulsive.

Hi in the first part of the given problem, this is silent night having a large number of turns of a conducting wire. One over the course, this is our resistance attached with the coin resistance R. Points air. And be it has been given that a bar magnet is moving up there from this sunday night such that the north pole of the bar magnet is going away. So there's a reason behind uh huh. E. M. F. Induced in this solenoid will be going away of this north wall and the magnetic reliance come out of the north pole. So as the north pole will move away from the left end of the solenoid, the magnetic flux passing through this left and will change due to the changing this magnetic flux. An IMF will be induced in the left phase of this solenoid. And as per lenses law, the nature of this E. M. F. Induced will be such that the magnetism induced due to that induced DNF will be such that it will oppose the reasons behind its induction. So as the reason is going away of the north pole, so the direction of current induced in the left end of the coil should be such that the left and the left face of the coil should start behaving like south pole so that it may attract the north pole of the bar magnet which is going away from it. So as north pole of the magnet is going away from the left face of the solenoid. So the direction of current induced in it should be such that this face need behave like south pole which who will at red the north pole mm going of it from me. So the current induced should be in clock wise direction in the left face. It should be with a clock wise direction needs to say like this, and if the direction of current in the left face is clockwise, so it will be getting down to point A. And then we'll pass through the resistance from A to be. So the direction of current induced in the left face. Yeah. Should be from a two B. This is the answer for the first part of this problem. For the first. Fever. No, for the second figure of the same problem again, here there is a core having our solenoid wound over it. The given manner like this, this is a cell and then there is a key attached with it. There is one more circuit here. The resistance are post terminal A. Is that India plane off paper and terminal B. Is out of the plane of paper and the circuit has been completed like this now as the current is switched on in this coil, the direction of current in the right face of the solenoid. Yes, clockwise. As seen from the right hand side means the right face of this solenoid will start behaving like a south pole and the south. The south polarity will be increasing as the current will be increasing due to which the magnetic field lines passing through the solenoid as well as the circuit. The secondary circuit will also be increasing. So an M. Phil being used in the secondary circuit also whose direction as for lenses law should be such that it should oppose the reason behind its induction and the reason is the south pole. The right face of the solenoid is becoming a south pole. So to oppose this south polarity, this left phase of the secondary circuit should also acquire. South polarity means the direction of currently unused and it should be clockwise. Or we can say in the resistance it should be from A. To B. Again. So if you write it on switching the current on in the solid night, it's right face will acquire south polarity S. The current in it is clockwise so to oppose it as per lenses law. Again, the left face off secondary circuit should also acquire south polarity. Or we can say the direction of current induced in it should be clockwise, meals in resistance are it should be from mm to be okay. In the 3rd part of the problem there is a circuit again, having a resistance are and terminals A. B. And this time this circuit has been kept. I just sent to us straight current carrying conductor carrying current do you left? And this current is reducing. It is decreasing in magnitude using right hand term rule. The direction of magnetic field due to this current and the magnetic field which is passing through this circuit is inward into the plane of paper. Magnetic field linked through the circuit is inward using right hand thumb rule and s. Current i is decreasing. We can say this inward magnetic field is decreasing due to which an M. F. And a current will be induced in the circuit whose direction should be such that to oppose the reason behind its induction. So the magnetically induced in the circuit should be inward increasing the induced magnetic field in the circuit inward increasing means it should start behaving life south pole as the magnetic field enters into the south pole, so the direction of currently in use and it should be clock vice. Or we can say the direction of quarantine. The resistance should be wrong mm To be again as the current when it is clockwise, The current in the resistance will be moving from a two beat. Thank you.

Hi. In the first part of the given problem, this is solid. Annoyed having a large number off terms off a conducting wire, one over the course. This is our resistance. Attached with the coil resistance are points a and beef. It has been given that a bar magnet is moving off the from this solenoid such that the North Pole off the bar magnet is going away. So as a reason behind nah e n f induced in this solenoid will be going away off this north pole and the magnetically lines come out off the North Pole. So as the North Pole will move away from the left end the solenoid the magnetic flux passing through this left and will change due to the changing this magnetic flux and NFL being used in the left face off this solid, annoyed and as per lenses law, the nature off this e m f induced will be such that the magnetism induced do toe that induce Vienna will be such that it will oppose the reasons behind its induction. So as the reason is going away off the North Pole. So the direction of current induced in the left end off the coil should be such that the left end, the left face off the coil should start behaving like South Pole so that it may attract the North Pole off the bar magnet, which is going away from it. So as North Pole off, the magnet is going away from the left face off the solid noid. So the direction off current induced in it should be such that this face meet behave like South board, which will attract the North Pole. Yeah, going on me from me. So the current induced should be in clock wise direction in the left face. It should be in the clock. Wise direction means to say like this. And if the direction of current in the left face is clockwise so it will be getting down to point a and then we'll pass through the resistance from a toe. Be so the direction off current induced in the left face should be from a to be. This is the answer for the first part off this problem for the first fever? No, For the second figure off the same problem again. Here there is accord. Having are solid, noid wound over it the given manner like this. This is a cell and then there is a key attached with it. There is one more circuit fear the resistance are who's terminal A is within the plane off paper and terminal B is out off the plane off paper and the circuit has been completed like this. Now, as the current is switched on in this school, the direction off current in the right face off the solenoid is clockwise as seen from the right hand site means the right faceoff. This solenoid will start behaving like ah, South Pole and the South. The South polarity will be increasing as the country of increasing due to which the magnetic field lines passing through this solar annoyed as well as this circuit, the secondary circuit will also be increasing, So an e m f will be induced in the secondary circuit. Also whose direction, as for lenses law should be such that it should oppose the reason behind its induction. And the reason is the South Pole, the right faceoff. The solenoid is becoming a South Pole so toe oppose this south polarity. This left face off the secondary circuit should also acquire South polarity means the direction of parenting news and it should be clockwise. Or we can say in the resistance it should be from a Toby again. So if you write it on switching the current own in the solid night, it's right face will acquire South Polarity s. The current in it is clockwise. So toe oppose it as per lenses law again, the left phase off secondary circuit should also acquire south polarity. Or we can see the direction off current induced in it should be clockwise. Meals in resistance are it should be from mm to be again in the third part of the problem, there is, ah, circuit again Having a resistance are and terminals a B. And this time this circuit has been kept. I just sent toe a straight current carrying conductors carrying current Do you left and this current is reducing. It is decreasing in magnitude using right hand thumb rule the direction off magnetic field due to this current and the magnetic field which is passing through this circuit, is inward into the plane off paper magnetic field linked through the circuit, his inward using right hand thumb rule, and as current I is decreasing. We can see this in world magnetic fields is decreasing due to which an E m f and a current will be induced in the circuit, whose direction should be such that toe Oppose the reason behind its induction. So the magnetically induced in the circuit should be inward, increasing the induced magnetic field in the circuit inward. Increasing means it should start behaving life South pole as the magnetic field enters into the South Pole. So the direction of current induced in it should be clock vice. Or we can say the direction of current in the resistance should be wrong. Mhm to be again as the current When it is clockwise, the current in the resistance will be moving from a to be thank you.


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