Question

A solenoid of length 0.5 m and diameter 0.6 cm consists of 1000 turns of fine wire carrying a current of $5.0 \times {10^{ - 3}}$ ampere. The magnetic field in Weber/ ${m^2}$ at the ends of the solenoid will be-
A. $7.8 \times {10^{ - 6}}$
B. $6.28 \times {10^{ - 6}}$
C. $3.14 \times {10^{ - 6}}$
D. $6.28 \times {10^{ - 5}}$

Answer 1

Hint: Magnetic field will be equal on both ends of the solenoid. So the magnetic field at the end of a long solenoid can be calculated by using the formula below, where B is the magnetic field, I is the current, N is the ratio of the number of turns over its length. Use this info to find the magnetic field.

Complete step by step answer:
Magnetic field at the end of a solenoid is $B = \dfrac{{{\mu _0}NI}}{2}$
We are given that a solenoid of length 0.5 m and diameter 0.6 cm consists of 1000 turns of fine wire carrying a current of $5.0 \times {10^{ - 3}}$ ampere.
We have to calculate the magnetic field in Weber/ ${m^2}$ at the ends of the solenoid.

Magnetic field is a vector quantity which describes the magnetic effect on moving electric charges and magnetic materials. A charge which is moving in a magnetic field experiences a force perpendicular to the magnetic field and to its velocity.
Magnetic field can be found using $B = \dfrac{{{\mu _0}NI}}{2}$
Where N is the ratio of no. of turns in the solenoid to the length of the solenoid
No. of turns is 1000 and the length is 0.4 m
$
  N = \dfrac{n}{L} \\
 \Rightarrow N = \dfrac{{1000}}{{0.4}} \\
 \Rightarrow N = 2500 \\
 $
${\mu _0}$ is the magnetic constant or the permeability of free space and its value is $4\pi \times {10^{ - 7}}H/m$
On substituting all the given and obtained values in the magnetic field formula, we get
 $
  B = \dfrac{{4 \times \pi \times {{10}^{ - 7}} \times 2500 \times 5 \times {{10}^{ - 3}}}}{2} \\
   \to B = \dfrac{{157079.63 \times {{10}^{ - 10}}}}{2} = 78539.81 \times {10^{ - 10}} \\
  \therefore B = 7.8 \times {10^{ - 6}}Weber/{m^2} \\
 $
Therefore, the correct option is Option A, $7.8 \times {10^{ - 6}}$

Note: Do not confuse the magnetic field of the solenoid at its centre with the magnetic field of the solenoid at its ends. The magnetic field at the centre of a solenoid is twice the magnetic field at the ends. One weber unit is the product of one Henry unit with one ampere unit.