physics

[THoang2005]

A .25 KG wooden cylinder .1 m in length and .2 m in diameter is wrapped longitudinally (the long way) with 10 loops of wire. the cylinder is then placed on an incline of angle (theta) where the plane of the wire is positioned parallel to the incline . the magnetic field is passing perpendicularly through the base of the incline. the magnetic field strength is .5 T . Find the current required to pass through the loops of wire to create a counter torque enough to keep the cylinder from rolling down the incline .


thanks

[THoang2005]

mass of cylinder = .25kg
length ----------=.1m
radius -------------= .2/2 = .1 m
10 loops
B = .5 T

F = I L B = m a
.25(9.8) = I (.1)(.5)
I = 49 A

torque = 10 ( 49 * area of cylinder * .5)
torque = 15. 39 Nm

am i rite ???

[THoang2005]

QT , please help
thanks

[Trung Quốc]

T(cylinder)=radius of a cylinder * Force of the cylinder
Force of the cylinder = graviational force * sin of theta

T = .1 * 9.8 * .25 * sin theta (1)

T(by current and magnetic field) = u x B
u = number of loops * I * (perpendicular area )
u = 10 * I * (.2 * .1 * sin theta)
T = .2* I * sin theta * .5 (2)

(1) and (2) have to equal so

.1 * 9.8 * .25 * sin theta = .2* I * sin theta * .5

.1 * 9.8 * .25 = .2* I * .5
I = .98 * .25 / .1
I = 2.45 A

Hope this right. Sorry I don't remember this very clearly so didn't help you before trying to see if anyone know it. :)

[THoang2005]

thanks QT

[yeudoi]

I can't remember electronic stuff. But QT's Torque formula is OK.