5. The sketch shows two parallel wires. The separation between the wires is L =

Question

5. The sketch shows two parallel wires. The separation between the wires is L = 16cm. Wire 1 carries a current of i1 = 5A and wire 2 carries a current of i2 = 20A in the direction shown in the figure. We place a third wire in the plane of these two, parallel to 1 and 2, such that the net force on wire 1 is zero and the net force on wire 2 is zero. A) Where should we place the third wire and what is the current 13 (both direction and magnitude!) carried by wire 3? B) What is the net force (both direction and magnitude) on wire 3? C) What is the net magnetic field (due to wire 1 and wire 2) at the location of wire 3?

Explanation / Answer

Let the wire 3 be placed x meters from the wire 1 to the left. And let the current flowing through it be 'i' flowing towards south(downwards)..

So, Force per unit length on wire 1 due to wire 3, F13 = u*(i)*(i1)/(2*pi*x) directed towards right <---- as wire 1 and wire 3 repel each other

Similarly, force on wire 1 due to wire 2, F12 = 8*(i1)*(i2)/(2*pi*L) directed towards left

So, net force on wire 1, Fnet1 = F13 - F12

For, Fnet1 = 0

So, F12 = F13

So,  u*(i)*(i1)/(2*pi*x) =  u*(i1)*(i2)/(2*pi*L)

So, i/x = i2/L

So, x = (i/i2)*L

So, i = x*i2/L --------(1)

Now,

net force on wire 2,

Fnet = F21 - F23 = 0

So, F21 = F23

So,  u*(i1)(i2)/(2*pi*L) = u*(i2)*i/(2*pi*(L+x))

So, i1/L = i/(L+x)

So, i = (L+x)*i1/L ------ (2)

Equating (1) and (2), we get,

x*i2/L = (L+x)*i1/L

So, x*(20)/(16) = (16+x)*5/(16)

So, x = 5.33 cm <------answer

So, i = 5.33*20/(16) = 6.67 A <------answer

So, wire 3 must be 5.33 cm to the left of wire 1 and the current is 6.67 A towards down <-----answer

b)

net force on wire 3,

Fnet3 = u*(i1)*i/(2*pi*x) - u*(i2)*i/(2*pi*(L+x)).

So, Fnet3 = 1.26*10^-6*6.67*(5/(2*pi*0.0533) - 20/(2*pi*(0.16+0.0533)))

= 5.88*10^-8 N/m towards left<----answer

c)

Net magnetic field, Bnet = Fnet3*i = 5.88*10^-8*(6.67) = 3.92*10^-7 T <-----answer

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