calculating magnetic fields (1 Viewer)

[waltssillyhat]

"A 30 cm long solenoid with a 2 cm diameter is to produce a 0.2 T magnetic field in its centre. If the maximum current is 8 A, how many turns must the solenoid have? (Assume that each coil is circular rather than helical.)"

I know that I have to use the formula B = (μNI)/L, but I don't understand where to use the given of the diameter. Do I use it to determine how much current the solenoid can handle? If so, how would I approach doing this?

 

[coolcat6778]

"A 30 cm long solenoid with a 2 cm diameter is to produce a 0.2 T magnetic field in its centre. If the maximum current is 8 A, how many turns must the solenoid have? (Assume that each coil is circular rather than helical.)"

I know that I have to use the formula B = (μNI)/L, but I don't understand where to use the given of the diameter. Do I use it to determine how much current the solenoid can handle? If so, how would I approach doing this?

I think u just use the formula B = (μNI)/L

 

[cheesynooby]

"A 30 cm long solenoid with a 2 cm diameter is to produce a 0.2 T magnetic field in its centre. If the maximum current is 8 A, how many turns must the solenoid have? (Assume that each coil is circular rather than helical.)"

I know that I have to use the formula B = (μNI)/L, but I don't understand where to use the given of the diameter. Do I use it to determine how much current the solenoid can handle? If so, how would I approach doing this?

I think it's a red herring

 

[coolcat6778]

the diameter is irrelevant. The formula B = (μNI)/L is made for calculating stuff at the centre, which you are given the values for

 

[m_avi]

mod 4 seems scary :\

 

[coolcat6778]

mod 4 seems scary :\

only circuits and then you will never have to do them again in the hsc

 

[C2H6O]

mod 4 seems scary :\

you aint seen nothing yet

 

[coolcat6778]

you aint seen nothing yet

mod 6 is easy

 

[C2H6O]

mod 6 is easy

yeah but tbh yr 12 phys in general is a huge step up compared to yr11 (or maybe just my opinion)

 

[coolcat6778]

yeah but tbh yr 12 phys in general is a huge step up compared to yr11 (or maybe just my opinion)

not really, i think year 11 physics is more math heavy while year 12 is less math heavy because module 7 and 8 are memorization and rote learning heavy instead.

year 11 physics has collisions (which can even be in 2d) circuits and snell's law, which imo are the hardest parts of high school physics

the HSC exams would be more differentiating if the prelim content was tested in lieu of the garbage 8 + 5 marker module 8 or 7 history questions

 

[C2H6O]

not really, i think year 11 physics is more math heavy while year 12 is less math heavy because module 7 and 8 are memorization and rote learning heavy instead.

year 11 physics has collisions (which can even be in 2d) circuits and snell's law, which imo are the hardest parts of high school physics

the HSC exams would be more differentiating if the prelim content was tested in lieu of the garbage 8 + 5 marker module 8 or 7 history questions

yeah thats valid

 

[cheesynooby]

nahhhhhhhh

 

[m_avi]

nahhhhhhhh

ches disagrees :0

 

[wizzkids]

"A 30 cm long solenoid with a 2 cm diameter is to produce a 0.2 T magnetic field in its centre. If the maximum current is 8 A, how many turns must the solenoid have? (Assume that each coil is circular rather than helical.)"

I know that I have to use the formula B = (μNI)/L, but I don't understand where to use the given of the diameter. Do I use it to determine how much current the solenoid can handle? If so, how would I approach doing this?

The magnetic field strength inside a solenoid is not significantly affected by the diameter.
This piece of information is a distractor.
The most important parameters that determine the magnetic field strength inside a solenoid are:

• Number of turns per unit length
• The current

However, if the question had asked how much magnetic flux is created inside the solenoid, that is where you would need to calculate the cross-sectional area of the solenoid and multiply by the field strength, and then you would obviously need the diameter.