magnet through solenoid q - help (1 Viewer)

[anonymoushehe]

how is it b and what difference does attaching wires at the ends of the solenoid do

 

[barnyard]

as the magnet falls, there is a changing magnetic field in the solenoid.
by faraday's law, an emf is induced by the changing flux.
this emf drives a current with a magnetic field which, by lenz's law, is in such a direction so as to oppose the change in flux that gave rise to its creation.
importantly, the emf can only drive a current if there is a closed loop, so only in A and B, due to the wire at the ends of the solenoid.
the magnet takes the longest when this opposing magnetic force slowing its fall is greatest, which occurs when there are more turns in the solenoid, ie B.

 

[anonymoushehe]

as the magnet falls, there is a changing magnetic field in the solenoid.
by faraday's law, an emf is induced by the changing flux.
this emf drives a current with a magnetic field which, by lenz's law, is in such a direction so as to oppose the change in flux that gave rise to its creation.
importantly, the emf can only drive a current if there is a closed loop, so only in A and B, due to the wire at the ends of the solenoid.
the magnet takes the longest when this opposing magnetic force slowing its fall is greatest, which occurs when there are more turns in the solenoid, ie B.

omg yeah so true thanks

 

[anonymoushehe]

as the magnet falls, there is a changing magnetic field in the solenoid.
by faraday's law, an emf is induced by the changing flux.
this emf drives a current with a magnetic field which, by lenz's law, is in such a direction so as to oppose the change in flux that gave rise to its creation.
importantly, the emf can only drive a current if there is a closed loop, so only in A and B, due to the wire at the ends of the solenoid.
the magnet takes the longest when this opposing magnetic force slowing its fall is greatest, which occurs when there are more turns in the solenoid, ie B.

wait do you mind if you also check if the answers are wrong; solutions say anticlockwise for the current direction (idk what they mean by that), i assumed current would from A to B, so is that the same thing?

 

[Tony Stark]

wait do you mind if you also check if the answers are wrong; solutions say anticlockwise for the current direction (idk what they mean by that), i assumed current would from A to B, so is that the same thing?

View attachment 47340

I'm confused...in such a coil, isn't there an equal and opposite force at either ends of the coil due to the magnets (motor effect)? To balance the mass, a force upwards would have to be produced equal to the weight of the mass, however wouldn't there already be another force downwards thus mitigating the balanced nature of the previous forces? Sorry if it doesn't anwer the question, but I just want to know

 

[quarkkkk]

the current will go from b to a (hence anticlockwise as the current flows from d -> c -> b -> a)
as the weight force at p goes into the page, you will need a motor effect force out of the page to balance the mass in side CD (hence ensuring net force is 0)
thus, you will do right hand rule (f=ilbsin(theta)) to get that the current flows from d to c on side CD
thus current will flow anticlockwise !!

 

[anonymoushehe]

the current will go from b to a (hence anticlockwise as the current flows from d -> c -> b -> a)
as the weight force at p goes into the page, you will need a motor effect force out of the page to balance the mass in side CD (hence ensuring net force is 0)
thus, you will do right hand rule (f=ilbsin(theta)) to get that the current flows from d to c on side CD
thus current will flow anticlockwise !!

im getting the current as A to B wouldnt the fingers point from right to left, and then you would want the palm to point outwards,so the thumb ends up pointing n the direction from A to B??

 

[anonymoushehe]

I'm confused...in such a coil, isn't there an equal and opposite force at either ends of the coil due to the magnets (motor effect)? To balance the mass, a force upwards would have to be produced equal to the weight of the mass, however wouldn't there already be another force downwards thus mitigating the balanced nature of the previous forces? Sorry if it doesn't anwer the question, but I just want to know

Yeah I see where you are coming from, I was thinking that the torque produced by the forces on each side of the coil had to be equal to the torque produced by the weight force of the mass???

 

[quarkkkk]

im getting the current as A to B wouldnt the fingers point from right to left, and then you would want the palm to point outwards,so the thumb ends up pointing n the direction from A to B??

you are using right hand rule correctly in that you are getting a current up the page
however as the mass is attached to side CD, thus when applying right hand rule, you are are looking at the current flowing on side CD instead of side AB.
thus your current will be flowing up the page from d to c
the current will thus go from d -> c -> b -> a

take note that current flowing in side ab will be in the opposite direction as current flowing in side cd as current flows in a loop !!

 

[anonymoushehe]

you are using right hand rule correctly in that you are getting a current up the page
however as the mass is attached to side CD, thus when applying right hand rule, you are are looking at the current flowing on side CD instead of side AB.
thus your current will be flowing up the page from d to c
the current will thus go from d -> c -> b -> a

take note that current flowing in side ab will be in the opposite direction as current flowing in side cd as current flows in a loop !!

wait omg so true i just realised so smart

 

[anonymoushehe]

you are using right hand rule correctly in that you are getting a current up the page
however as the mass is attached to side CD, thus when applying right hand rule, you are are looking at the current flowing on side CD instead of side AB.
thus your current will be flowing up the page from d to c
the current will thus go from d -> c -> b -> a

take note that current flowing in side ab will be in the opposite direction as current flowing in side cd as current flows in a loop !!

while you are here why is the answer for this D pls

 

[cheesynooby]

while you are here why is the answer for this D pls
View attachment 47341

XY moves left: emf is in a specific direction to oppose this change in flux.
XY moves right: emf must reverse direction to oppose this again.
XY moves left (back to original direction): emf must go back in the original direction
this is only observed on A or D

emf = -N∆flux/t
so A seems unlikely as it suggests that it's moving at high speed to the left and then accelerates instantly towards the right
(as emf can be thought of as proportional to the velocity of XY)
so it's D

 

[anonymoushehe]

i beg someone explain how did they get P to Q

my thought process was

increase on scale --> force on magnet by rod is downwards ---> force on rod by magnet must be upwards then using Newtons third law of motion.

then use right hand rule
- palm facing up, thumb point from north to south and then the current moved along the direction from Q to P?

 

[barnyard]

i think you're getting confused with the forces or how a spring balance works which is fair bc that diagram sucks
the coil/rod is what is attached to the spring balance, so when the reading increased, that meant the coil/rod went DOWN, meaning the force ON THE coil/rod BY THE magnet is DOWN.
then just do RHR as you did before

 

[anonymoushehe]

i think you're getting confused with the forces or how a spring balance works which is fair bc that diagram sucks
the coil/rod is what is attached to the spring balance, so when the reading increased, that meant the coil/rod went DOWN, meaning the force ON THE coil/rod BY THE magnet is DOWN.
then just do RHR as you did before

omg wait i didnt realise it was a spring balance thought the rod was on the magnet resting oopsies tyyy

 

[barnyard]

omg wait i didnt realise it was a spring balance thought the rod was on the magnet resting oopsies tyyy

yeaa it's okay, my biggest mistake with physics has ALWAYS been misreading or misinterpreting the question so ive learnt to try and take it slower and absorb the question instead of rushing into it and making silly mistakes

 

[anonymoushehe]

yeaa it's okay, my biggest mistake with physics has ALWAYS been misreading or misinterpreting the question so ive learnt to try and take it slower and absorb the question instead of rushing into it and making silly mistakes

also how did they get 2.3 A for the last one, i assumed that the maximum possible force was 5 N in the question so I used F = nBIL --> 5 = 100(0.3)(I)(5x10^-2) --> 3.30 A - where am i messing up

 

[barnyard]

the coil already has a weight force of 1.6N so it can only increase by 5-1.6=3.4N
F=nBIL -> 3.4=100(0.3)(I)(5x10^-2) -> I=2.27

 

[anonymoushehe]

the coil already has a weight force of 1.6N so it can only increase by 5-1.6=3.4N
F=nBIL -> 3.4=100(0.3)(I)(5x10^-2) -> I=2.27

omg you so smart