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Hi there!
We can use Ampère's Law:
[tex]\oint B \cdot dl = \mu_0 i_{encl}[/tex]
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:
[tex]\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[/tex]
Since AD and BC are perpendicular, and since:
[tex]\oint B \cdot dl = B \cdot L = BLcos\phi[/tex]
[tex]BLcos(90) = 0[/tex]
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:
[tex]\oint B \cdot dl = \int\limits^C_D {B} \, dl[/tex]
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:
[tex]B L = \mu_0 Ni[/tex]
N = # of loops per length multiplied by the length, so:
[tex]BL = \mu_0 nL i \\\\B = \mu_0ni[/tex]
Plug in the given values and solve. Remember to convert # of loops to # of loops per unit length.
[tex]B = \mu_0 (100/0.1)(2) = (4\pi *10^{-7})(1000)(2) = \boxed{0.00251 T}[/tex]