Please do all 4 parts I am confused The picture below shows

Please do all 4 parts I am confused!!!

The picture below shows a cross-sectional view through three long straight current-carrying wires, which are each carrying current perpendicular to the page (or the screen). In this problem, use these values: the distance a = 40.0 cm; the current I = 5.50 amps.

(a) Calculate the net force per unit length acting on the wire that carries a current of 3I out of the page, at x = +2a. Use a plus sign if the force is directed to the right, and a minus sign if the force is directed to the left.

(b) Calculate the net magnetic field at x = +a. Use a plus sign if the field is directed up, and a minus sign if the field is directed down.

(c) Calculate the magnitude of the net magnetic field at at a distance of a directly above the wire located at x = 0.

(d) Find the distance from x = 0 to the point closest to x = 0 where the net magnetic field is zero. ___cm

Solution

We need to keep following things into consideration while solving the given problem:

1.) The magnetic field due to an infinitely long wire at a distance a is given as I/ 2a. The direction of this field can be obtained by the right hand rule where we keep the thumb along the current and wrap the wire with our fingers. The curl of the fingers give the direction of the magnetic field.

2.) The force acting on a current carrying wire due to magnetic field B is given as BIL where L is the length. However, in case of infinitely long wire the force per unit length would be BI, where B is field and I is the current in the wire for which force is to be determined. The direction of this force is given by the cross product of the current\'s direction and the the field direction.

We will use the above points to solve the given problems as follows:

a.) For net force, we will determine the net field and its direction.

Field due to I at origin =  I/ 4a and direction would be upwards.

Field due to 3I at x = -2a would be: -3I/ 8a.

Net field = -I/ 8a and the direction as indicated by minus sign would be downwards.

hence the force would be 3I²/ 8a = 4 x 10^-7 x 3 x 5.5*5.5 / 8*0.4 = 113.4375 x 10^-7 N and the direction would towards ther right. [Cross product of current direction and field]

b.) For x = +a

Field due to 3I at -2a:

B1 = -3I/ 6a

Field due to I at origin: B2 = I/ 2a

Field due to 3I at x = +2a:

B3 = - 3I/ 2a

Net B = -3I/ 6a - 3I/ 2a + I/ 2a = (3I - 9I - 3I)/ 6a = -3I / 2a

B = -3I / 2a = 3 x 4 x 10^-7 x 5.5 / 2 x 0.4 = - 82.5 x 10^-7 T

c.) For a point at distance a above the wire at origin,

the distance from 3I at -2a and +2a would be: sqrt(5a)

The fields due to the two wires at -2a and +2a would be at 90 degrees to each other. the net field due to these two would hence point towards the origin and each will have same magnitude (By symmetry)

Hence B1 = -3I2 / 25a

B due to I at the origin would be along -x axis and have magnitude of -I/ 2a

Hnece the net B = (I / 2a)sqrt[18/5 + 1] = (I / 2a)(sqrt23/5) = (4 x 10^-7 x 5.5 / 0.8)(2.14476)

Therfore, B= 58.981 x 10^-7 T

d.) Consider a point at distance x along the -x axis for x< -2a.

Here, the fields due to I and 3I on positive side would add up and the field due to 3I on -x side would counter it.

Hence / 2 [3I/(x-2a) - I/x - 3I / (x+2a)] = 0

or, I/ 2[ (3x(x+2a) - x^2 + 4a^2 -3x( x - 2a))/ x(x^2 - 4a^2)] = 0

For the above expression to be zero, we must have:

(3x(x+2a) - x^2 + 4a^2 -3x( x - 2a) = 0

or,x^2 - 12ax - 4a^2 = 0

or x = 12a +/- asqrt160] / 2 = 4.9298 metres to the left of origin