Two straight parallel wires carry currents in opposite directions as shown in th

Question

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I2 = 10.9 A. Point A is the midpoint between the wires. The total distance between the wires is d = 12.9 cm. Point C is 5.28 cm to the right of the wire carrying current I2. Current I1 is adjusted so that the magnetic field at C is zero. Calculate the value of the current I1.

Please answer A,B,and C

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I2-10.9 A. Point A is the midpoint between the wires. The total distance between the wires is d = 12.9 cm. Point C is 5.28 cm to the right of the wire carrying current 12. Current I1 is adjusted so that the magnetic field at C is zero. Calculate the value of the current I1 Suomit Anawer Tries 0/20 Calculate the magnitude of the magnetic field at point A Submit AnswerTries 0/20 What is the force between two 1.45 m long segments of the wires?

Explanation / Answer

a)

distance of point from first wire through current i1 = r1 = 12.9+5.28

r1 = 18.18 cm = 0.1818m
distance of point from first wire through current i2 = r2 = 0.0528 m
i2 = 10.9 A

i1 = i2 (r1 / r2) = 10.9 X 0.1818 / 0.0528

i1 = 10.9 X 3.44

i1 = 37.53 amps

the value of current i1 = 37.53 amps
b)

magnetic field at point A
B1 = 2 k (i1/r)
B2 = 2 k (i2/r)
where

r = d /2 = 12.9 /2

r = 6.45 cm = 0.0645m

k = o / 4

= 10-7 Weber / Ampere - meter
total magnetic field

B = B1 + B2

B = 2 k (i1/r) + 2 k (i2/r)

B = (2 k / r) (i1 + i2 )

B = (2 X 10-7 / 0.0645) (10.9 + 37.53 )

B = (2 X 10-7 / 0.0645) (48.43)

B = 96.86 X 10-7 / 0.0645

B = 1501 .70 X 10-7

B = 1.501 X 10-4 T

magnetic field at point A is B = 1.501 X 10-4 T
c)

the force for unit length = F / L

= 2 k i1 i2 / d

but given d = 12.9 cm

d = 0 .129 cm
and

given length of the wire segments is

L = 1.45 m
F = 2 k i1 i2 L / d
F = 2 X 10-7 X 37.53 X 10.9 X 1.45 / 0.129

F = 1186. 3233 X10-7 / 0.129

F = 9196.30 X10-7 N

or

F = 9.196 X 10-4 N

9.196 X 10-4 N is the force between two 1.45 m

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