MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR:
The equation `vecF = q(vecv xx vecB)` can be used to find the force on a
current carrying conductor placed inside the magnetic
field `B`. Consider an element of length `dl` of an infinitely
long wire carrying a current ; placed inside a field `B` as
shown in the figure.
The free electrons in the wire drift with a speed `v_d`
known as the drift speed of the electrons in the
direction opposite to the current. The relation between
the current ; and the drift speed `v_d` is.
`i = jA = n e(v_d)A`
Here A is the area of cross section and n is the number
of free electrons per unit volume. The number of
electrons in an elemental volume A dl is nA dl. Thus the
magnetic force on a wire of length dl will be
`dvecF = (nAdl)(-evecv_d xx vecB)`
where q = -e is the charge on electron
If we denote length dl along the direction of the current
by dl , the above equation becomes
`dvecF = nAe v_d vecdl xx vecB`
Using (i),
`dvecF = ivecdl xx vecB` ............(4.1)
The quantity `(i vecdl)` is called a current element.
If a straight wire of length l carrying a current i is placed
in a uniform magnetic field `vecB`, the force acting on it is
given by
`vecF = ivec(dl) xx vecB` ............................(4.2)
Points to remember regarding expression (4.2)
(1) Magnitude of `vecF` is , F = `ilB sintheta`, here `theta` is the angle
between `vecl` and `vecB` .F is zero for
`e = 0^0`. or `180^0` and maximum for `theta = 90^0`
(2) Here `vecl` is the vector that points in the direction of
current i and has magnitude equal to the length.
(3) The above expression applies only to a straight
segment of wire in a uniform magnetic field.
(4) For the magnetic force on an arbitrarily shaped wire
segment, let us consider the
magnetic force exerted on a small segment of vector
length vecdl
`dvecF ==i(dvecl xx vecB )`...........................(4.3)
To calculate the total force F acting on the wire shown
in the figure above, we integrate Eqn. 4.3 over the
length of the wire.
`F = iint_A^D(dvecl xx vecB)` .................(4.4)
Two special cases arise for the Eqn. 4.4
Case 1 . A curved wire ACD as shown in the Fig. (a)
Carries a current I and is located in a uniform magnetic field `vecB` . Because the field is uniform, we can take `vecb`
outside the integral in Eqn. 4.4 and we obtain.
`vecF = (iint_A^D (dvecl)) xx vecB_0` ....................(4.5)
But the quantity `int_A^D(dvecl)` represents the vector sum of all
length elements from A to D. From the polygon law of
vector addition the sum equals the vector `vecl` directed
from A to D. Thus,
`vecF = ivecl xx vecB`....................(4.6)
Or we can write, `F_(ACD) = F_(AD) = i(vecAD xx vecB)`
Case 2. An arbitrarily shaped closed loop carrying a
current I is placed in uniform magnetic field as shown
in Fig. (b). We can again express the force acting on the
loop in the form of Eqn.4.5, but this time we must take
the vector sum of the length element `dvecl_0`
Over the entire loop.
`vecF = (iointdvecl) xx vecB`
Because the set of length elements forms a closed
polygon, the vector sum must be zero.
`vecF = 0`
Thus, the net magnetic force acting on any closed
current loop in a uniform magnetic field is zero.
(5) The direction of F can be given by Fleming's left
hand rule . According to this rule, the forefinger, the
middle Finger and the thumb or the left hand are
stretched in such a way that they are mutually
perpendicular to each other. If middle finger shows the
direction of current (or) and forefinger shows the
direction of magnetic lleld `(vecB)`, then the thumb will give
the direction of magnetic force (F).
current carrying conductor placed inside the magnetic
field `B`. Consider an element of length `dl` of an infinitely
long wire carrying a current ; placed inside a field `B` as
shown in the figure.
The free electrons in the wire drift with a speed `v_d`
known as the drift speed of the electrons in the
direction opposite to the current. The relation between
the current ; and the drift speed `v_d` is.
`i = jA = n e(v_d)A`
Here A is the area of cross section and n is the number
of free electrons per unit volume. The number of
electrons in an elemental volume A dl is nA dl. Thus the
magnetic force on a wire of length dl will be
`dvecF = (nAdl)(-evecv_d xx vecB)`
where q = -e is the charge on electron
If we denote length dl along the direction of the current
by dl , the above equation becomes
`dvecF = nAe v_d vecdl xx vecB`
Using (i),
`dvecF = ivecdl xx vecB` ............(4.1)
The quantity `(i vecdl)` is called a current element.
If a straight wire of length l carrying a current i is placed
in a uniform magnetic field `vecB`, the force acting on it is
given by
`vecF = ivec(dl) xx vecB` ............................(4.2)
Points to remember regarding expression (4.2)
(1) Magnitude of `vecF` is , F = `ilB sintheta`, here `theta` is the angle
between `vecl` and `vecB` .F is zero for
`e = 0^0`. or `180^0` and maximum for `theta = 90^0`
(2) Here `vecl` is the vector that points in the direction of
current i and has magnitude equal to the length.
(3) The above expression applies only to a straight
segment of wire in a uniform magnetic field.
(4) For the magnetic force on an arbitrarily shaped wire
segment, let us consider the
magnetic force exerted on a small segment of vector
length vecdl
`dvecF ==i(dvecl xx vecB )`...........................(4.3)
To calculate the total force F acting on the wire shown
in the figure above, we integrate Eqn. 4.3 over the
length of the wire.
`F = iint_A^D(dvecl xx vecB)` .................(4.4)
Two special cases arise for the Eqn. 4.4
Case 1 . A curved wire ACD as shown in the Fig. (a)
Carries a current I and is located in a uniform magnetic field `vecB` . Because the field is uniform, we can take `vecb`
outside the integral in Eqn. 4.4 and we obtain.
`vecF = (iint_A^D (dvecl)) xx vecB_0` ....................(4.5)
But the quantity `int_A^D(dvecl)` represents the vector sum of all
length elements from A to D. From the polygon law of
vector addition the sum equals the vector `vecl` directed
from A to D. Thus,
`vecF = ivecl xx vecB`....................(4.6)
Or we can write, `F_(ACD) = F_(AD) = i(vecAD xx vecB)`
Case 2. An arbitrarily shaped closed loop carrying a
current I is placed in uniform magnetic field as shown
in Fig. (b). We can again express the force acting on the
loop in the form of Eqn.4.5, but this time we must take
the vector sum of the length element `dvecl_0`
Over the entire loop.
`vecF = (iointdvecl) xx vecB`
Because the set of length elements forms a closed
polygon, the vector sum must be zero.
`vecF = 0`
Thus, the net magnetic force acting on any closed
current loop in a uniform magnetic field is zero.
(5) The direction of F can be given by Fleming's left
hand rule . According to this rule, the forefinger, the
middle Finger and the thumb or the left hand are
stretched in such a way that they are mutually
perpendicular to each other. If middle finger shows the
direction of current (or) and forefinger shows the
direction of magnetic lleld `(vecB)`, then the thumb will give
the direction of magnetic force (F).