A straight conductor of length 5cm carries current of 15A.The
conductor experience a magnetic force of 4.5X10-3N
when it is piaced in a magnetic field of 0.9T.Whatangle
the conductor makes with magnetic field.
Answers
Answer:
Because charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself.
A diagram showing a circuit with current I running through it. One section of the wire passes between the north and south poles of a magnet with a diameter l. Magnetic field B is oriented toward the right, from the north to the south pole of the magnet, across the wire. The current runs out of the page. The force on the wire is directed up. An illustration of the right hand rule 1 shows the thumb pointing out of the page in the direction of the current, the fingers pointing right in the direction of B, and the F vector pointing up and away from the palm.
Figure 1. The magnetic field exerts a force on a current-carrying wire in a direction given by the right hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire, since typical currents consist of very large numbers of moving charges.
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
Ø=3.84°
Explanation:
Length of conductor (l) = 5cm = 0.05m
Current in conductor (i) = 1.5 A
Magnetic force (B) = 0.9 T
Angle(ø) = ?
We have,
F = Bilsinø
Or, 4.5×10^-3 = 0.9×0.05×1.5×sinø
Or, sinø = 0.067
.
. . Ø = 3.84°