A short current element dl= 10.500 mm2ne carries a current of 5.40 A in the same direction as dl. Point P is located atr= (-0.730 m)i^+ (0.390 m)j^. Use unit vectors to express the magnetic field at P produced by this current element.

Short Answer

Expert verified

Using the unit vectors to express the magnetic field at the produced current element is

dBx,dBy,dBz=1.85.10-10,0,3.47.10-10T

Step by step solution

01

Unit vector

Using the Biot-Savart law

μ04πidl×r^r2

Now turning into unit vector, we get:

-0.73,0,0.390.732+0.39

02

Magnetic field at point p

Now after plugging the values in the equation, we get:

μ04π5.40,0.0005,0×-0.88,0,0.470.832

Now computing the cross product by finding the determinant of the matrix we get:

ijk00.00050-0.8800.47

Putting the cross products back into the Biot-savart equation we get

μ04π5.42.3.10-4,0,4.4.10-40.832

Therefore, the magnetic field at point P is

dBx,dBy,dBz=1.85.10-10,0,3.47.10-10T

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