Question

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

• A. 2
• B. 1/2
• C. 1
• D. 3/2

Answer 1

Answer: (A)

The magnetic field at some internal point on the axis of a solenoid is given by
$B=\frac{{\mu }_{0}nI}{2}\left(\cos {\theta }_1-\cos {\theta }_2\right)$
Magnetic field at the centre of a long solenoid is given by setting ${\theta }_1=0^{\circ }$ and ${\theta }_2=\pi$.
$\Rightarrow B_1=\frac{{\mu }_{0}nI}{2}\left(\cos 0^{\circ }-\cos \pi \right)$
$=\frac{{\mu }_{0}nI}{2}[1-(-1)]$
$={\mu }_{0}nI...(i)$
Similarly, the magnetic field at the end of a long solenoid is given by setting ${\theta }_1=\frac{\pi }{2}$ and ${\theta }_2=\tau$
$\Rightarrow B_2=\frac{{\mu }_{0}nI}{2}\left(\cos \frac{\pi }{2}-\cos \pi \right)$
$=\frac{{\mu }_{0}nI}{2}[0-(-1)]$
$=\frac{{\mu }_{0}nI}{2}...(ii)$
From Eq. (i) and Eq. (ii), we get
$\frac{B_1}{B_2}=\frac{{\mu }_{0}nI}{{\mu }_{0}nI}\times 2=2$