thecsstudent Posted April 24, 2016 Report Share Posted April 24, 2016 Why is the answer to this A? I know the induced current should create a magnetic field that opposes the one that created it but if the answer is A wouldn't it be in the same direction below the wire? Also in this one why is the answer D?Isn't the force from the left magnet to the right and the force from the right magnet to the right as well? Reply Link to post Share on other sites More sharing options...
indimpi Posted April 24, 2016 Report Share Posted April 24, 2016 Hi, For the first one you can just apply the right hand rule, and see that the force on the negative charges (i.e. electrons) is directed towards end Q, and so end P will have fewer electrons. For the second one, the proton is at rest. The magnetic force on a charged particle in rest is zero (F=qvB), and so it should be D. 2 Reply Link to post Share on other sites More sharing options...
Vioh Posted April 24, 2016 Report Share Posted April 24, 2016 For question 23, you need to use the right-hand rule for induction by magnetic force. This rule states that if you align your thumb in the direction of the velocity (downward in this case), and the rest of your fingers in the direction of the magnetic field (which points from North to South, i.e. to the right in this case), then the palm of your right hand will point along the direction of the force acting on the positive charge (out of the page), and the back of the palm will point along the direction of the force acting on the negative charge (into the page). Since the electrons are negatively charged, they should flow into the page, in the direction from P to Q. Hence, P will have fewer electrons compared to Q. 3 hours ago, thecsstudent said: Why is the answer to this A? I know the induced current should create a magnetic field that opposes the one that created it but if the answer is A wouldn't it be in the same direction below the wire? It also seems to me here that you are a bit confused about Lenz's law (which states that the magnetic field created by the induced current will oppose the change the magnetic field acting on the wire). Here you must remember that there are 2 types of electromagnetic induction (i.e. 2 ways that you can induce current in a wire). The first way is induction by changing flux, which is applied to loops of wires that are moving relative to a changing magnetic field. If this is the case that you are dealing with, then you must use Lenz's law to deduce the direction of the induced current. The second way is induction by magnetic force, which is applied to straight wires (similar to question 23). In this case, you must use the right-hand rule to deduce the direction of the induced current. For question 19, as indimpi has already mentioned, you should use the formula F=qvB. And since the velocity of the proton is v=0, thus the force acting on it is also 0. There is indeed a force acting on the left bar magnet from the right bar magnet, and there is a force acting on the right bar magnet from the left bar magnet. However, the questions asks for the magnetic force acting on the proton, so you need to use F=qvB formula. 3 Reply Link to post Share on other sites More sharing options...
thecsstudent Posted April 24, 2016 Author Report Share Posted April 24, 2016 3 hours ago, Vioh said: For question 23, you need to use the right-hand rule for induction by magnetic force. This rule states that if you align your thumb in the direction of the velocity (downward in this case), and the rest of your fingers in the direction of the magnetic field (which points from North to South, i.e. to the right in this case), then the palm of your right hand will point along the direction of the force acting on the positive charge (out of the page), and the back of the palm will point along the direction of the force acting on the negative charge (into the page). Since the electrons are negatively charged, they should flow into the page, in the direction from P to Q. Hence, P will have fewer electrons compared to Q. It also seems to me here that you are a bit confused about Lenz's law (which states that the magnetic field created by the induced current will oppose the change the magnetic field acting on the wire). Here you must remember that there are 2 types of electromagnetic induction (i.e. 2 ways that you can induce current in a wire). The first way is induction by changing flux, which is applied to loops of wires that are moving relative to a changing magnetic field. If this is the case that you are dealing with, then you must use Lenz's law to deduce the direction of the induced current. The second way is induction by magnetic force, which is applied to straight wires (similar to question 23). In this case, you must use the right-hand rule to deduce the direction of the induced current. For question 19, as indimpi has already mentioned, you should use the formula F=qvB. And since the velocity of the proton is v=0, thus the force acting on it is also 0. There is indeed a force acting on the left bar magnet from the right bar magnet, and there is a force acting on the right bar magnet from the left bar magnet. However, the questions asks for the magnetic force acting on the proton, so you need to use F=qvB formula. Thanks so much about this I actually have been confused with Lenz's law for a while and this cleared it up for me! Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.