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Motion of a Charged Particle inside Electric Field
FORCE ON A CURRENT CARRYING CONDUCTOR PLACED IN A MAGNETIC FEILD
Consider a straight conductor PQ of length / placed along Y-axis. Suppose that a uniform magnetic field of strength \(\overrightarrow B\) acts along Z-axis i.e. perpendicular to the length of the conductor [Fig.]. Further, suppose that current I flows through the conductor from its end P to Q, so that the vector \(\overrightarrow I l \) (current element vector) is along OY
H.A. Lorentz was the first to suggest that the electric current in metals is due to the flow of electrons. In a conductor, current flows due to drift of the electrons. If the conductor has area of cross-section A and the number of free electrons per unit volume equal to n drifting with velocity va, then the current flowing through the conductor is given by
\( I = nAev_d \)
Multiplying both sides by l, we have
\( Il = nAlev_d \)
The direction of flow of the conventional current I is opposite to that of the drift velocity \(\overrightarrow {v_d } \) of the electrons. In other words, \( \overrightarrow I l \) and \( \overrightarrow {v_d } \) have opposite directions and likewise the above equation may be written as
\( \overrightarrow I l = - nAle\overrightarrow {v_d } \) .........(1)
The magnetic Lorentz force on an electron having charge -e and moving with velocity \(\vec v_d\) inside the magnetic field of strength \(\vec B\) is given by
\( \overrightarrow f = - e\left( {\overrightarrow {v_d } \times \overrightarrow B } \right)\) .........(2)
Each of the free electron inside the conductor experiences force ƒ under the effect of magnetic field. If N is the total number of electrons in the conductor, then force on the current carrying conductor placed inside the magnetic field is given by
\( \overrightarrow F = n\overrightarrow f \) ........(3)
Now, the total number of free electrons in the conductor,
N=n x volume of conductor = n A l
Substituting for N and \(\vec f\) in equation (3), the force on the conductor,
\( \overrightarrow F = nAl\left( { - e\left( {\overrightarrow {v_d } \times \overrightarrow B } \right)} \right) = \left( { - nAle\overrightarrow {v_d } } \right) \times \overrightarrow B \)
Using equation (1), we have
\( \overrightarrow F = \overrightarrow I l \times \overrightarrow B \) ........(4) -
Motion of a Charged Particle inside Electric Field
Magnitude of force.
From equation (4), the magnitude of the force on the current carrying conductor is given by
F= \( BIl\) sin\(\theta \) ,...(5)
where is the angle between the direction of magnetic field and the direction of flow of current.
(i) If \(\theta\)=0° or 180° i.e. sin \(\theta\) = 0, then F=\( BIl\) (0)=0
Thus, if the current carrying conductor is placed parallel to the direction of magnetic field, it does not experience any force.
(ii) If \(\theta\)=90°,then
F=\( BIl\) sin 90°=\(BIl\) (maximum)
In other words, a current carrying conductor experiences maximum force in a magnetic field, when it is placed perpendicular to the direction of magnetic field. -
Motion of a Charged Particle Electric Field
Direction of force.
If \(\vec Il\) and \(\vec B\) are oriented along OY and OZ respectively as shown in Fig. , then the equation (4) tells that the force \( \overrightarrow F = \overrightarrow I l \times \overrightarrow B \) on the conductor will act along OX.
In practice, to find the direction of force on a current carrying conductor, Fleming's left hand rule is used.
FLEMING’S LEFT HAND RULE
It states that if the fore-finger, central finger and the thumb of left hand are stretched mutually perpendicular to each other, such that the fore-finger points in the direction of magnetic field (\(\vec B\)) and the central finger in the direction of current (I), then the thumb points in the direction of the force (\(\vec F\)) experienced by the conductor. The rule has been illustrated in Fig.
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Motion of a Charged Particle inside Electric Field
FORCE BETWEEN TWO INFINITELY LONG PARALLEL CURRENT CONDUCTORS
Consider two infinitely long conductors X1Y1, and X2Y2 placed parallel to each other at a distance r apart. The currents I, and I are flowing through the two conductors ,X1Y1 and X2Y2 in the same direction as shown in Fig. .
Let us find the force experienced per unit length by the current carrying conductor X1Y1 due to the magnetic field produced by the current carrying conductor X2Y2 Magnetic field at point P (on the conductor X1Y1) due to curernt I, flowing through the infinitely long conductor X2Y2 is given by
\( B_2 = \frac{{\mu _0 }} {{4\pi }}\frac{{2I_2 }} {r}\) ------(6)
According to the right hand thumb rule, the direction of the magnetic field B2 at point P is perpendicular to the plane of paper and in inward direction. Now, the conductor X1Y1 carrying current I1 lies in the magnetic field B2 produced by the conductor X2Y2 .Since \( F = BIl\), the force experienced by the unit length of the conductor X1Y1 due to magnetic field B2 is given by
\( F = B_2 \times \left( {I_1 \times 1} \right) = \frac{{\mu _0 }} {{4\pi }}\frac{{2I_2 }} {r} \times I_1 \)
\( F = \frac{{\mu _0 }} {{4\pi }}\frac{{2I_1 I_2 }} {r}\)............(7)
Applying Fleming's left hand rule, it follows that the force F on the conductor X1Y1 acts in the plane of the paper and towards left. If we proceed in a similar manner, then it can be proved that the conductor X2Y2 experiences an equal force in the plane of the paper but towards right.
a) When the currents in both the wires is in the same direction, then mutual force between the wires is attractive. When the currents in the wires are in opposite directions, The mutual force between the wires is repulsive.
b) Mutual force between the wires is \( F = \frac{{\mu _0 }} {{4\pi }}\frac{{i_1 i_2 }} {r}l\)
Force per unit length on each wire is \( \frac{F} {l} = \frac{{\mu _0 }} {{4\pi }}\frac{{i_1 i_2 }} {r}\)
Note: If i=i2 1A, r = 1m then F= \( 2 \times 10^{ - 7} Nm^{ - 1} \)
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Motion of a Charged Particle inside Electric Field
Right hand thumb rule.
If we grasp the conductor in the palm of the right hand so at the thumb points in the direction of the flow of current, then the direction in which the fingers curl, gives the direction of magnetic field lines. This rule is illustrated in Fig.
It follows that magnetic field is in the form of concentric circles, whose centres lie on the straight conductor.
c) Right hand palm rule:
Stretch the fingers and thumb of right hand at right
angles to each other. If the fingers point in the direction of field \(\vec B\) and thumb in the direction of current i, the plam will point in the direction of force or motion.
d) The magnetic force on the current element is non centra.
e) The net magnetic force on a current loop in a uniform magnetic field is always zero. Here different points of the loop may experience elemental force due to which the loop may be under tension or may experience a torque.
f) A loop of flexible conducting wire of length '' lies in a magnetic field induction B perpendicular to the plane of the loop.