electromagnetism--a2 Physics

Questions:

1) A solenoid uses in magnetic resonance imaging (MRI) is would from a niobiumtitanium superconducting wire 2.0mm in diameter, with adjacent turns sperated by an insulating layer of negligible thickness. (The solenoid is 2.4m long and 95cm in diameter). Calculate the current in the superconducting wires necessary to produce a magnetic field of flux density 1.5T at the centre of the solenoid.

2) The magnetic flux density at a distance of 10mm from a long straight wire carrying a current of 10A is 0.20mT. What is the magnetic flux density
(a) at a distance of 6.0mm, with the current still 10A
(b) at a distance of 10mm, if the current is 3.0A

3) A horizontal conductor of length 50mm carrying a current of 3.0A lies at right angles to a horizontal magnetic field of flux density 0.50T. Draw a diagram to show the direction of the current, the field and the force.

Thanks in advance! I am really stuck on the above...have a test on it in 10 days...so all help is appreciated! :biggrin:

Reply 1

SinghFello

2

We've only just started this, but ill have a crack at it.

1. B(solanoid flux)=k(=4[pi]x10^-7])xnxI(current)

n=N/L
24cm/2mm = 120 Turns
Circumference of solanoid = [Pi]D = 2.98m
2.98m x 120 = 357.6m (length of wire)
n= 0.335

1.5=4[pi]x10^-7x0.335xI

so I get current= 3.5x10^6A (but this seems too high?)

2. Flux of straight wire = k(same as above)xI/2[pi]r(distance from wire)

not sure about this one. k cant equal 4[pi]x10^-7 or B wouldnt be 0.2mT(assuming m = milli not mega?)

So using the figures, k = 0.0125
using this new k
a = 0.12mT
b = 0.0596mT

3. f= 0.075N (f=bil)
I think you use fleming left hand rule to show that force is going right, the magnetic field is going up, and the current is going downwards(on the y axis)

Im still really bad at these, so they will probably be incorrect!

Reply 2

habosh

13

devesh254

Questions:

1) A solenoid uses in magnetic resonance imaging (MRI) is would from a niobiumtitanium superconducting wire 2.0mm in diameter, with adjacent turns sperated by an insulating layer of negligible thickness. (The solenoid is 2.4m long and 95cm in diameter). Calculate the current in the superconducting wires necessary to produce a magnetic field of flux density 1.5T at the centre of the solenoid.

well if I understood right the solenoid is made from a wire that is 2mm in diameter so to find number of turns 2.4/2*10^-3=1200 as thickness od the insulating layers is negligible.
so 1.5/(4(pi)*10^-7*(1200/2.4))=2.39KA

2) The magnetic flux density at a distance of 10mm from a long straight wire carrying a current of 10A is 0.20mT. What is the magnetic flux density
(a) at a distance of 6.0mm, with the current still 10A

use B=(muI)/(2(pi)r
so B=3.33mT

(b) at a distance of 10mm, if the current is 3.0A

B=.06mT

(3) A horizontal conductor of length 50mm carrying a current of 3.0A lies at right angles to a horizontal magnetic field of flux density 0.50T. Draw a diagram to show the direction of the current, the field and the force.

Thanks in advance! I am really stuck on the above...have a test on it in 10 days...so all help is appreciated! :biggrin:

the conductor should be vertical magentic field horizantal if current is upwords (towards you) magnetic field is to the right and force is upwordsuse flemings left hand rule

Reply 3

habosh

13

SinghFello

We've only just started this, but ill have a crack at it.

1. B(solanoid flux)=k(=4[pi]x10^-7])xnxI(current)

n=N/L
24cm/2mm = 120 Turns
Circumference of solanoid = [Pi]D = 2.98m
2.98m x 120 = 357.6m (length of wire)
n= 0.335

1.5=4[pi]x10^-7x0.335xI

so I get current= 3.5x10^6A (but this seems too high?)

(ill edit the rest in)

wrong
n is number of turns per unit length you have to find number of turns and devide them over2.4
and it's 2 mm you have to convert both to unit length of meter
in a solenoid the magnetic flux is independent of the cross sectional area if the length of the solenoid is largh compared to it\s diameter so you don't use it here
and 2.4 m is 240 cm :wink: