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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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The magnetic field at some internal point on the axis of a solenoid is given by
$B=\frac{\mu_0 n I}{2}\left(\cos \theta_1-\cos \theta_2\right)$
Magnetic field at the centre of a long solenoid is given by setting $\theta_1=0^{\circ}$ and $\theta_2=\pi$.
$\Rightarrow \quad B_1=\frac{\mu_0 n I}{2}\left(\cos 0^{\circ}-\cos \pi\right)=\frac{\mu_0 n I}{2}[1-(-1)]$
$=\mu_0 n I$ $\ldots(i)$
Similarly, the magnetic field at the end of a long solenoid is given by setting $\theta_1=\frac{\pi}{2}$ and $\theta_2=\pi$.
$\Rightarrow \quad B_2=\frac{\mu_0 n I}{2}\left(\cos \frac{\pi}{2}-\cos \pi\right)=\frac{\mu_0 n I}{2}[0-(-1)]$ $\ldots(ii)$
From Eq. (i) and Eq. (ii), we get
$\frac{B_1}{B_2}=\frac{\mu_0 n I}{\mu_0 n I} \times 2=2$
$B=\frac{\mu_0 n I}{2}\left(\cos \theta_1-\cos \theta_2\right)$
Magnetic field at the centre of a long solenoid is given by setting $\theta_1=0^{\circ}$ and $\theta_2=\pi$.
$\Rightarrow \quad B_1=\frac{\mu_0 n I}{2}\left(\cos 0^{\circ}-\cos \pi\right)=\frac{\mu_0 n I}{2}[1-(-1)]$
$=\mu_0 n I$ $\ldots(i)$
Similarly, the magnetic field at the end of a long solenoid is given by setting $\theta_1=\frac{\pi}{2}$ and $\theta_2=\pi$.
$\Rightarrow \quad B_2=\frac{\mu_0 n I}{2}\left(\cos \frac{\pi}{2}-\cos \pi\right)=\frac{\mu_0 n I}{2}[0-(-1)]$ $\ldots(ii)$
From Eq. (i) and Eq. (ii), we get
$\frac{B_1}{B_2}=\frac{\mu_0 n I}{\mu_0 n I} \times 2=2$
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