An electric current flowing through a conductor produces a magnetic field.
The field so produced exerts a force on a magnet placed in the vicinity of the
conductor.
i) The figure shows the current carrying conductor is placed between two
magnets with different poles. Find the direction of force produced in
the sliding conductor.
a) P b) Q c) R d) S
ii) Which of the following will affect the magnitude of the force acting on a
current carrying conductor placed in a magnetic field.
I. Length of the conductor
II. Magnitude of the current
III. Diameter of the conductor
IV. Angle between the current and magnetic field
a) I and II only b) II and III only
c) II, III and IV only d) I, II and IV only
iii) An electron enters a magnetic field at right angles to it, as shown in
figure. The direction of force acting on the electron will be
a) to the right b) to the left c) out of the page d) into the page
iv) A positively – charged particle (alpha particle) projected towards west is
deflected towards north by a magnetic field. The direction of magnetic
field is
a) towards south b) towards east c) downward d) upward
v) A positively – charged particle (alpha particle) projected towards East is
deflected towards south by a magnetic field. The direction of magnetic
field is
a) towards south b) towards east c) downward d) upward
Answers
Answer:
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, as in Figure 2.
Illustration of the right hand rule 1 showing the thumb pointing right in the direction of current I, the fingers pointing into the page with magnetic field B, and the force directed up, away from the palm.
Figure 2. The force on a current-carrying wire in a magnetic field is F = IlB sin θ. Its direction is given by RHR-1.
EXAMPLE 1. CALCULATING MAGNETIC FORCE ON A CURRENT-CARRYING WIRE: A STRONG MAGNETIC FIELD
Calculate the force on the wire shown in Figure 1, given B = 1.50 T, l = 5.00 cm, and I = 20.0 A.