Question

A long solenoid (diameter = 5.0 cm) is wound with 960 turns per meter of thin wire through which a current of 300 mA is maintained. A wire carrying 12 A is inserted along the axis of the solenoid. What is the magnitude of the magnetic field at a point 2.0 cm from the axis?

Answer 1

Answer: The magnetic field B= 4.82×10^-4T

Explanation:

There are two sources of magnetic field; the magnetic field due to solenoid and magnetic field due to wire. Their sum will give the total magnetic field

The magnetic field due to solenoid is constant at any point while the magnetic field due to wire varies with 1/radius.

Since they are perpendicular to each other;

Using Pythagoras theorem

B^2 =(B1)^2+B2)^2

B1 is the magnetic field due to solenoid and B2 is the magnetic field due to wire

B1= (μNI1)/L

B2=μI2/2πr

Where N/L is the of turns per meter = 960m^-1

I1 is the current in the solenoid =300mA=0.3A

I2 is the current in the wire = 12A

r is the distance from the axis of solenoid = 2cm= 0.02m

μ is the magnetic constant= 4π×10^-7Tm/A

Magnetic field due to solenoid B1=(4π×10^-7×960×0.3)

B1= (3.6191×10^-4 )

Magnetic field due to wire

B2 = (4π×10^-7×12)/2π × 0.02

B2 = 1.2×10^-4

B^2 =( 3.16191×10^-4)^2 × (1.2×10^-4)^2

B=3.8×10^-4T

Therefore the magnetic field is 0.38mT