Obtain am expression for magnetic
force on a wire carrying a comment
Answers
Explanation:
We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd is given by F = qvdB sin θ. Taking B to be uniform over a length of wire l and zero elsewhere, the total magnetic force on the wire is then F = (qvdB sin θ)(N), where N is the number of charge carriers in the section of wire of length l. Now, N = nV, where n is the number of charge carriers per unit volume and V is the volume of wire in the field. Noting that V = Al, where A is the cross-sectional area of the wire, then the force on the wire is F = (qvdB sin θ) (nAl). Gathering terms,
F
=
(
n
q
A
v
d
)
l
B
sin
θ
.
Because nqAvd = I (see Current),
F
=
I
l
B
sin
θ
is the equation for magnetic force on a length l of wire carrying a current I in a uniform magnetic field B, as shown in Figure 2. If we divide both sides of this expression by l, we find that the magnetic force per unit length of wire in a uniform field is
F
l
=
I
B
sin
θ
. The direction of this force is given by RHR-1, with the thumb in the direction of the current I. Then, with the fingers in the direction of B, a perpendicular to the palm points in the direction of F,