How Is the Magnetic Field Calculated Near a Current-Carrying Wire?

[SimonZ]

Homework Statement

A straight wire carrying a current of 42 A lies along the axis of a 6.6 cm-diameter solenoid. The solenoid is 70 cm long and has 250 turns carrying a current of 6.0 A.
Estimate the magnitude of the magnetic field 4.2 cm from the wire.

Homework Equations

magnetic field due to a straight current I
B = mu_0*I/(2*pi*r)

The Attempt at a Solution

Note 4.2 cm > radius 3.3 cm, so the point is outside the solenoid, the magnetic field is only due to the straight current
use I = 42 A, r = 4.2 cm, get B = 0.20 mT
Anything wrong?

 

[rock.freak667]

Note 4.2 cm > radius 3.3 cm, so the point is outside the solenoid, the magnetic field is only due to the straight current
use I = 42 A, r = 4.2 cm, get B = 0.20 mT
Anything wrong?

The solenoid is 70 cm long and has 250 turns carrying a current of 6.0 A.

The solenoid carries a current, so it has a magnetic field. For a solenoid

$$B=\mu_0nI \ where \ n=\frac{No. \ of \ turns}{Length \ of \ solenoid}$$


So you'd need to find the resultant mag. field of the solenoid and the straight conductor.

 

[SimonZ]

B = μo(N/l) * I = μonI
is only valid for magnetic field INSIDE the solenoid.
The field OUTSIDE the solenoid is zero.
So the field is only due to the straight current