At New York City, the earth\'s magnetic field has a vertical component of Magnit

Question

At New York City, the earth's magnetic field has a vertical component of

Magnitude ____N

Direction ________

A 45-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 67 A and experiences a magnetic force of 0.15 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 54.0° with respect to the wire.
T

The x, y, and z components of a magnetic field are Bx = 0.12 T, By = 0.14 T, and Bz = 0.17 T. A 25-cm wire is oriented along the z axis and carries a current of 5.3 A. What is the magnitude of the magnetic force that acts on this wire?
N

Explanation / Answer

1. In general, the force on a current, I, of length, L, in a magnetic field ,B,
is the vector equation;
F = (IL) x B
Let (i,j,k) be the directions (East, North, Vertical Up) this will form a right handed system; curl East into North with right hand and thumb points UP.
You are given;
current element = (IL)i
magnetic field = (By)j - (Bz)k
then the force is , F = (IL)i X [(By)j - (Bz)k]
= (IL)(By)(i X j) - (IL)Bz(i X k)
= (IL)(By)k + (IL)(Bz)j
These are the z & y components of F. To get the magnitude square, add , take sqroot;
F^2 = (IL)^2[(Bz)^2 + (By)^2)
SO finally,
F = ILSqRt[(Bz)^2 + (By)^2]
= (34)(28)SqRt[27.04 x 10^-10 + 3.24 x 10^-10]=(34)*(28)*(5.50*10^-5)=0.0523N

2.F=I*[L × H], where I = 67A, |F| = 0.15N, |L|=45m, |H| is magnetic field to be found;
|F| = I* |[L × H]| = I* |L| *|H| *sin(54°);
hence |H| = |F| /{I*|L|*sin(54°)} =(0.15)/(67*45*sin(54))=6.14*10^-5T

3.X component of magnetic force,Fx=ILBx
=5.3x0.25x0.12=0.159 N
Y component of magnetic force,Fy=ILBy
=5.3x0.25x0.14
=0.1855 N
Z component of magnetic force,Fz=ILBz
=5.3x0.25x0.17
=0.2252 N
Net magnetic force,F= (Fx^2 + Fy^2 + Fz^2)
= ( (0.159)^2 + (0.1855)^2 + (0.2252)^2 )
=0.3322N

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