Question

A solenoid 1.5 m long and 4.0 cm in diameter possesses 10 turns/ cm. A current of 5 A is flowing through it. Then the magnetic induction (i) inside and (ii) at one end on the axis of solenoid are respectively
$\begin{align}
  & \text{A}\text{. }2\pi \times {{10}^{-3}}T,\pi \times {{10}^{-3}}T \\
 & \text{B}\text{. }3\pi \times {{10}^{-3}}T,1.5\pi \times {{10}^{-3}}T \\
 & \text{C}\text{. }0.5\pi \times {{10}^{3}}T,4\pi \times {{10}^{3}}T \\
 & \text{D}\text{. }4\pi \times {{10}^{3}}T,2\pi \times {{10}^{3}} \\
\end{align}$

Answer 1

Hint: We are given the length diameter and the number of turns per centimeter of a solenoid. We need to find the magnetic induction inside the coil and at the end of the coil when a current of 5A is passed through it. We know the equations to find magnetic induction inside and at the ends of a solenoid. By substituting the values in it we get the solution.

Formula Used:
Magnetic induction inside the solenoid,
$B={{\mu }_{0}}nI$
Magnetic induction at the end on the axis of the solenoid,
$B=\dfrac{{{\mu }_{0}}nI}{2}$

Complete step-by-step answer:
In the question we have a solenoid. Length and diameter of the solenoid is given to us.
Length of the solenoid = 1.5m
Diameter of the solenoid = 4.0 cm
Number of turns per centimeter of the solenoid is given to be 10.
Therefore number of turns per meter of the solenoid,
$\begin{align}
  & n=\dfrac{10}{{{10}^{-2}}}\times 1.5 \\
 & n=1500\text{turns/m} \\
\end{align}$
It is said that a current of 5.0A is passing through the solenoid.
First we need to calculate the magnetic induction inside the solenoid.
We know that the magnetic field inside a solenoid is given by the equation,
$B={{\mu }_{0}}nI$, were ‘B’ is the magnetic induction, ‘${{\mu }_{0}}$’ is the permeability constant or permeability of free space, ‘n’ is the number of turns per unit length and ‘I’ is the current flowing through the solenoid.
We know that permeability of free space,
${{\mu }_{0}}=4\pi \times {{10}^{-7}}$
Earlier we found the number of turns per unit length,
n = 1500 turns/m
And we know current flowing through the solenoid,
$I=5A$
Therefore we have the magnetic induction inside the solenoid,
$\begin{align}
  & B=4\pi \times {{10}^{7}}\times 1500\times 5 \\
 & B=3\pi \times {{10}^{-3}}T \\
\end{align}$
Now we need to find the magnetic induction at one end of the solenoid.
We know that the magnetic induction at one end on the axis of the solenoid is given by,
$B=\dfrac{{{\mu }_{0}}nI}{2}$
We know the value of ${{\mu }_{0}}nI$ from the previous calculation.
${{\mu }_{0}}nI=3\pi \times {{10}^{-3}}$
Therefore magnetic induction at one end on the axis of the solenoid is,
$\begin{align}
  & B=\dfrac{3\pi \times {{10}^{-3}}}{2} \\
 & B=1.5\pi \times {{10}^{-3}}T \\
\end{align}$
Now we have,
Magnetic induction inside the solenoid $=3\pi \times {{10}^{-3}}T$
Magnetic induction at one end on the axis of the solenoid $=1.5\pi \times {{10}^{-3}}T$
Hence the correct answer is option B.

Note: A solenoid is a coil of wire that acts as a magnet when current is passed through it.
We know that the magnetic induction is given by the formula,
$B={{\mu }_{0}}nI$
In this equation ‘n’ is not the number of turns. It is the number of turns per unit length. And length should be in meters not in centimeters.
In the question we are given the number of turns per centimeter, hence we convert that into the number of turns per meter.