Question

A 10-cm long solenoid has 100 turns and a radius of 5 cm. If it carries a current of 2 A, What is the magnetic field B inside the solenoid

Answer 1

Hi there!

We can use Ampère's Law:

$\oint B \cdot dl = \mu_0 i_{encl}$

B = Magnetic field strength (B)
dl = differential length element (m)
μ₀ = Permeability of free space (T/Am)

Since this is a closed-loop integral, we must integrate over a closed loop. We can integrate over a rectangular-enclosed area of the rim of the solenoid - ABCD - where AD and BC are perpendicular to the solenoid.

Thus, the magnetic field is equivalent to:
$\oint B \cdot dl = \int\limits^A_B {B} \, dl + \int\limits^B_C {B} \, dl + \int\limits^C_D {B} \, dl + \int\limits^D_A {B} \, dl$

Since AD and BC are perpendicular, and since:
$\oint B \cdot dl = B \cdot L = BLcos\phi$

$BLcos(90) = 0$

If perpendicular to the field, the equation equals 0.

Additionally, since AB is outside of the solenoid, there is no magnetic field present, so B = 0. The only integral we integrate now is:
$\oint B \cdot dl = \int\limits^C_D {B} \, dl$

Which is horizontal and inside the solenoid. Let the distance between C and D be 'L', and the enclosed current is equivalent to the number of loops multiplied by the current:

$B L = \mu_0 Ni$

N = # of loops per length multiplied by the length, so:
$BL = \mu_0 nL i \\\\B = \mu_0ni$

Plug in the given values and solve. Remember to convert # of loops to # of loops per unit length.

$B = \mu_0 (100/0.1)(2) = (4\pi *10^{-7})(1000)(2) = \boxed{0.00251 T}$