- An electron is moving at 6 x 107 m/s
at a distance of 50 mm from a long straight wire carrying a current of
40 A. Find the magnitude and direction of the force on the electron
when it is moving parallel to the wire
(a) in the same direction as the current and (b) in the opposite
direction to the current.
Bwire =
(µ0 I)/(2πr)
Direction is at
right angles to charge motion.
F = q v Bwire
= (q v µ0
I)/(2π r) = (1.6 x 10-19
x 6 x 107 x 4π
x 10-7
x 40)/( 2π x 5 x10-2)
= 1.5 x 10-15 N
Magnitude in (a)
and (b) are the same.
Directions : (a)
away from wire, (b) towards wire.
- A charge of +2.0 x 10-6 C is moving at
1.0 x 103 m/s at a distance of 120 mm away from a straight
wire
carrying a current of 4 A. Find the magnitude and direction of the
force on the charge when it is moving
perpendicular to the wire (a) towards the wire and (b)away from the
wire.
Bwire =
(µ0 I)/(2π
r)
Direction is at
right angles to charge motion.
F = q v Bwire
= (q v µ0
I)/(2πr) = (2.0 x 10-6
x 103 x 4π x 10-7
x 4)/( 2π x 0.12) = 1.3 x
10-8 N
Magnitude in (a)
and (b) are the same.
Directions : (a)
opposite the current, (b) same
direction as current.
- A horizontal wire 600 mm long whose mass is 4 g
is to be supported magnetically against the force of
gravity. The current in the wire is 12 A and goes from South to North.
Find the magnitude and direction
of the magnetic field of least magnitude that will support the wire.
For equilibrium FB
= mg where FB
= I L B assuming B and wire at right angles.
Therefore, mg = I
L B
B = (mg)/(I L) =
(4 x 10-3 x 9.8)/(12 x
0.6) = 5.44 x 10-3 T
Direction of
magnetic field, by right hand rule, is
from East to West.
- The parallel wires in a lamp cord are 2 mm apart.
What is the force per metre between them when the cord
is used to supply power to a 120 V, 200 W light bulb ?
Power = I V
I = P/V = 200/120
= 1.67 A
F/L = (µ0
I1 I2)/(2πd) = (4πx 10-7
x 1.67 x 1.67)/(2π x 2 x
10-3) =
2.79 x 10-4 N/m
Force is repulsive
since currents are in opposite
directions.