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Q.
The ratio of the magnetic field inside a solenoid at an axial point well inside and at an axial end point is
TS EAMCET 2020
Solution:
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 $\theta_{2}=\pi$.
$\Rightarrow B_{1}=\frac{\mu_{0} n I}{2}\left(\cos 0^{\circ}-\cos \pi\right)$
$=\frac{\mu_{0} n I}{2}[1-(-1)]$
$=\mu_{0} n I ... (i)$
Similarly, the magnetic field at the end of a long solenoid is given by setting $\theta_{1}=\frac{\pi}{2}$ and $\theta_{2}=\tau$
$\Rightarrow B_{2}=\frac{\mu_{0} n I}{2}\left(\cos \frac{\pi}{2}-\cos \pi\right)$
$=\frac{\mu_{0} n I}{2}[0-(-1)]$
$=\frac{\mu_{0} n I}{2} ... (ii)$
From Eq. (i) and Eq. (ii), we get
$\frac{B_{1}}{B_{2}}=\frac{\mu_{0} n I}{\mu_{0} n I} \times 2=2$