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Main Menu Contents Grades » Homework #7-Magnetic Field & Magnetic Force » Timer Notes e valuate .,Feedback Print Course Contents » A wire having a linear mass density of 1.00 g/cm is placed on a horizontal surface that has a coefficient of friction of 0.211. The wire carries a current of 1.69 A toward the east and slides horizontally to the north. What is the magnitude of the smallest magnetic field that enables the wire to move in this fashion?.20678 T Submit Answer Incorrect. Tries 2/8 Previous Tries This discussion is closed Send Feedback

Explanation / Answer

let magnetic field be applied at angle theta(measured from the normal to the surface) and Fb be the force due to magnetic field.

Magnetic force Fb=BIL
the problem asks for B
then
B=Fb/IL
here 'I' is the current ................'L' is the length of the wire.

Our system is in full equilibrium
so
Fy=0
n+FbCos-W=0 ........[ 'n' is the normal force due to surface............'W' is the weight of the wire ...W = mg ]
recall that frictional force........... f=n
and n=f/

n+FbCos-W=0
n=W-FbCos
f/ = W-FbCos
f=(W-FbCos)........................................... eq1

Fx=0
FbSin-f=0
FbSin=f ....................................eq2


equate eq1 and eq2
(W-FbCos) = FbSin
W-FbCos= FbSin/
W=FbCos + FbSin/
W=Fb(Cos + Sin/)
W=Fb(Sin+Cos)/
Fb=W/(Sin+Cos)


taking the derivative of above equation, we have to minimize it
dFb/d = W(Cos-Sin)/(Sin+Cos)^2
0=W(Cos-Sin)/(Sin+Cos)^2
simplifying we have
Cos-Sin=0
Cos=Sin
tan=1/
=0.211
=ArcTan(1/0.211)
=78.08degrees

from
B=Fb/IL
B=(m/L)g/I(Sin+Cos)
B=0.211(0.1kg/m)(9.8m/s^2)/[1.69A(Sin78.08 + 0.211*cos(78.08)]
B=0.202T

Therefore
B=0.202T at 78.08deg

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