Respuesta :
Answer:
Concluding aspect of the question is
1. The position of wire 3
2. The magnitude and direction of current in wire 3
Explanation:
Given that,
Current in wire 1, i1 =1.2A upward
Current in wire 2, i2=5A downward
Distance between wire 1 and 2 =20cm=0.2m
d1→2=0.2m
μo=4π×10^-7
The force per unit length on wire 1 by wire 2 is given as
F/l =μo•i1•i2/2πr
F/l= (4π×10^ -7×1.2×5)/(2π×0.2)
F/l=6×10^-6N/m
The force per unit length on wire 2 by wire 1 is the same as above
F/l=6×10^-6N/m
So, this is the magnitude of force each of wire1 and wire2 exerted on wire3
Also, since the current on wire2 is higher than wire1, the wire3 should be closer to wire1 than wire2 to balance the field.
Let wire3 be x m from wire 1, left of wire1
Then, wire 2 will be x+0.2 from wire3.
So, the force per unit length exerted on wire 3 by wire1 is equal to force per unit length exerted by wire3 by wire1
F/l (1→3) = F/l (2→3)
μo•i1•i3/2πx = μo•i2•i3/2π(x+0.2)
Now from the equation above
μo, i3, 2π, cancels out
Then, we are left with
i1/x = i2/(x+0.2)
1.2/x = 5/(x+0.2)
Cross multiply
1.2(x+0.2)=5x
1.2x+0.24=5x
0.24=5x-1.2x
0.24=3.8x
x=0.24/3.8
x=0.0632m
So wire 3 is at position 0.0632m from wire 1 and it is at 0.2632m from wire 2.
b. To calculate current
Using
F/l=μo•i1•i3/2πx
Since, F/l=6×10^-6N/m
Then,
6×10^-6 = μo•i1•i3/2πx
6×10^-6=4π×10^-7×1.2×i3/(2π×0.0632)
Cross multiply
6×10^-6×2π0.0632=4π×10^-7×1.2×i3
2.383×10^-6=4π×10^-7×1.2×i3
Then,
i3= (2.383×10^-6)/(4π×10^-7×1.2)
i3= 1.58A and it's direction is downward.
The question is not complete and the complete question is;
Ask Your Teacher Three long wires (wire 1, wire 2, and wire 3) hang vertically. The distance between wire 1 and wire 2 is 20.0 cm. On the left, wire 1 carries an upward current of 1.20 A. To the right, wire 2 carries a downward current of 5.00 A. Wire 3 is located such that when it carries a certain current, each wire experiences no net force. Find the following;
(a) the position of wire 3 in cm
(b) the magnitude and direction of the current in wire 3
Answer:
A) Wire 3 is 6.32cm from wire 1
and 26.32 cm from wire 2.
B) Current in wire 3 =1.58 A ;It's direction is downwards.
Explanation:
The force (per unit length) on wire 1 by 2 is to the left and is given as;
F/L = (μo.•I1•I2)/2πr
Where;
The ratio F/L is the force per unit length between the two parallel currents
I1 and I2 are the 2 parallel currents
r is the distance separating the 2 parallel currents.
μo is permeability of free space and it has a constant value of 4π x 10^(-7) T.m/A
The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.
I1 = 1.2A
I2 = 5A
r = 20cm = 0.2m
Thus ;
F/L = (4π x 10^(-7) x 1.2 x 5)/(2 x π x 0.2) = 6 x 10^(-6) N/m
From third law, we now have the force on wire 2 by wire 1 to be 6 x 10^(-6) N/m and to the right.
Hence, the magnitude of the force on each wire by wire 3 must be 6 x 10^(-6) N/m
Since the current in wire 2 is greater than that in wire 1, then wire 3 must be closer to wire 1 than 2.
Thus, wire 3 must be to the left of wire 1
Let x be the distance from wire 3 to wire 1 and so;
distance from wire 3 to wire 2 = x + 0.20
Thus, to find the position of wire 3;
(μo x I1 x I3)/2πx = (μo x I3 x I2)/(2π(x + 0.20))
Some values will cancel out to get;
I1/x = I2/(x+0.20)
Since I1 = 1.2A and I2 = 5A
Thus,
1.2/x = 5/(x + 0.2)
5x = 1.2(x+0.2)
5x = 1.2x + 0.24
5x - 1.2x = 0.24
3.8x = 0.24
x = 0.24/3.8
x = 0.0632m = 6.32 cm
Wire 3 is 6.32cm from wire 1
From wire 2 = 0.0632 + 0.2 = 0.2632 or 26.32 cm
B) Now, (μo•I1•I3)/(2πx) = 6 x 10^(-6)
so, I3 = [6 x 10^(-6) x (2πx)]/(μo•I1)
Thus;
6 x 10^(-6) x (2π•0.0632)/(4π x 10^(-7) x 1.2) = 1.58 A and its direction is down
