Two long, straight, parallel wires, 10.0 cm apart, carry equal 4.00 A currents in the same direction, as shown in Fig. Find the magnitude and direction of the magnetic field at pointP3, 20.0 cm directly aboveP1.

Short Answer

Expert verified

The magnetic field at point is 7.54μTin left direction.

Step by step solution

01

Step 1:  The magnetic field due to wire at a point

The magnetic field due to wire at point is given by

B=μ0l2πr

Where, B is the magnetic field due to wire,μ0is the permeability of vaccum, l is the current through the wire and r is the distance from wire.

The total magnetic field at point due two wires

The magnetic field at point due two wires is given by vector summation of the magnetic field of two wires at that point.

BT=B1x+B2x

Where, BTis the total magnetic field, B1xandB2xare magnetic fields due to first wire and second wire.

02

Calculation of distance of point P3 from wire 1  and wire 2 .

Given : The current carried by both wire is l = 4.00A

The distance between two wires is 20.0cm

Using Pythagoras theorem

Hypotenuse2=(perpendicular)2+(Base)2r2=(20.0cm)2+(5.00cm)2r=400+25r=20.6cm

Where, r is the distance of point P3from wire 1 and wire 2

03

Calculation of the total magnetic field at point due two wires 

Using

BT=B1x+B2xBT=2μ0I2πrcosθ

Where , θ is the angle between of magnetic field vector and horizontal.

Now, putting the values of constants in above equation

BT=2μ0l2πrcosθi^BT=4×107×420.6×102×0.2020.6×102i^BT=(7.54μT)i^

Thus, the magnetic field at point is 7.45μTin left direction .

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