Question

Two long thin parallel wires 13.0 cm apart carry 25-A currents in the same direction.

Part A

Determine the magnitude of the magnetic field vector at a point 10.0 cm from one wire and 6.0 cm from the other (Figure 1) .

Express your answer using two significant figures. B=?

Part B

Determine the direction of the magnetic field vector at that point.

Express your answer using two significant figures.

Answer 1

Given that,

distance between the wires = d = 13 cm = 0.13 m ; Current = I = 25 A

(A) d1 = 6 cm = 0.06 m and d2

Let B1 be the magnetic field for 0.06 m distance and B2 be from 0.1 m distance.

B1 = o I / 2 pi d1 = 4 pi x 10^-7 x 25 / 2 pi 0.06 = 833.3 x 10^-7 = 8.33 x 10^-5 T

B2= o I / 2 pi d2 = 4 pi x 10^-7 x 25 / 2 pi 0.1 = 833.3 x 10^-7 = 5 x 10^-5 T

We need to know the x and y components of magnetic field for which we need to determine the angle. As from the given distances its clear that the triangle thus formed is not a right angle traingle. So, lets use the property of triangle to find the angle.

d1² = d2² + d² - 2 d2 d cos1

(0.06)² = (0.1)² + (0.13)² - 2 x 0.1 x 0.13 x cos1

Solving this we get, 1 = 26.34 deg

similarly for 2 ; d2² = d1² + d² - 2 d1 d cos2

(0.1)² = (0.06)² + (0.13)² - 2 x 0.06 x 0.13 x cos2

2 = 47.7 degrees

Now we can write for the net field as:

B = B1(-cos1 i + sin1 j) + B2 (cos2 i + sin2 j)

B = 8.33 x 10^-5 T (-cos(26.34) + sin(26.34) + 5 x 10^-5 T (cos (47.7) + sin (47.7)

B = (-3.75 x  10^-5 ) i  + (7.1 x  10^-5 ) j = sqrt [(-3.75 x  10^-5)² + (7.1 x  10^-5)² ]

B = 8.03 x 10^-5 T

Part(B)

Direction will be given by:

Tan = B2/B1 = 7.1/3.75 => = 62.2 degrees.

Hence, = 62.2 deg