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(A) The magnetic field at an axial point of a solenoid is given by $B = \frac{\mu_0 n I}{2}(\cos 	heta_1 - \cos 	heta_2)$.For a point well inside th

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$2$

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(A) The magnetic field at an axial point of a solenoid is given by $B = \frac{\mu_0 n I}{2}(\cos 	heta_1 - \cos 	heta_2)$.For a point well inside the solenoid (at the center), $	heta_1 = 0^{\circ}$ and $	heta_2 = 180^{\circ}$.Thus, $B_{	ext{center}} = \frac{\mu_0 n I}{2}(\cos 0^{\circ} - \cos 180^{\circ}) = \frac{\mu_0 n I}{2}(1 - (-1)) = \mu_0 n I$.For a point at the axial end of the solenoid, $	heta_1 = 90^{\circ}$ and $	heta_2 = 180^{\circ}$.Thus, $B_{	ext{end}} = \frac{\mu_0 n I}{2}(\cos 90^{\circ} - \cos 180^{\circ}) = \frac{\mu_0 n I}{2}(0 - (-1)) = \frac{\mu_0 n I}{2}$.The ratio of the magnetic field at the center to the end is $\frac{B_{	ext{center}}}{B_{	ext{end}}} = \frac{\mu_0 n I}{\frac{\mu_0 n I}{2}} = 2$.

The ratio of the magnetic field inside a solenoid at an axial point well inside and at an axial end point is

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