Question

A solenoid of length $0.4m$ and diameter $0.6m$ consists of a single layer of $1000$ turns of fine wire carrying a current of $5 \times {10^{ - 3}}\,A$. Calculate the magnetic field on the axis at the middle and at the end of the solenoid:
A. $8.7 \times {10^{ - 6}}\,T$ , $6.28 \times {10^{ - 6}}\,T$
B. $6.28 \times {10^{ - 6}}\,T$ , $8.7 \times {10^{ - 6}}\,T$
C. $5.7 \times {10^{ - 6}}\,T$ , $6.28 \times {10^{ - 6}}\,T$
D. $8.7 \times {10^{ - 6}}\,T$ , $8.28 \times {10^{ - 6}}\,T$

Answer 1

Hint: A solenoid is defined as a coil that will act as a magnet when the current will pass through it. When the current passes through the coil, it will induce a magnetic field around the coil. Here, we will use the Biot-Savart’s law to calculate the magnetic field on the solenoid.

Complete step by step answer:
Consider a solenoid of length $0.4m$ and diameter $0.6m$ that consists of $1000$ turns of fine wire that is carrying a current of $5 \times {10^{ - 3}}\,A$.
Therefore, the length of solenoid, $ = 0.4\,m$
Also, the diameter of solenoid, $ = 0.6\,m$
Number of turns in solenoid, $n = 1000$
Current in solenoid, $I = 5 \times {10^{ - 3}}\,A$
Now, the magnetic field at a point on the axis of a solenoid is given by
$B = \dfrac{{{\mu _0}nI}}{2}\left( {\sin \alpha + \sin \beta } \right)$
Now, we know that at the center of the solenoid, the angle of the magnetic field will be equal, therefore
$\alpha = \beta $
Therefore, the magnetic field at the center of the solenoid is given by
$B = \dfrac{{{\mu _0}nI}}{2}\left( {\sin \alpha + \sin \alpha } \right)$
$ \Rightarrow \,B = \dfrac{{{\mu _0}nI}}{2} \times 2\sin \alpha $
$ \Rightarrow \,B = {\mu _0}nI\sin \alpha $
Now, substituting the values in the above equation, we get
$ \Rightarrow \,B = 4\pi \times {10^{ - 7}} \times \dfrac{{1000}}{{0.4}} \times \left( {5 \times {{10}^{ - 3}}} \right) \times \left( {\dfrac{{0.2}}{{\sqrt {{{\left( {0.3} \right)}^2} + {{\left( {0.2} \right)}^2}} }}} \right)$
$ \Rightarrow \,B = 4\pi \times \dfrac{{10000}}{4} \times {10^{ - 10}} \times 5 \times \left( {\dfrac{{0.2}}{{0.0036}}} \right)$
$ \Rightarrow \,B = 5\pi \times {10^{ - 6}} \times 0.555$
$ \Rightarrow \,B = 2.775\pi \times {10^{ - 6}}$
$ \Rightarrow \,B = 2.775 \times \dfrac{{22}}{7} \times {10^{ - 6}}$
$ \therefore \,B = 8.73 \times {10^{ - 6}}\,Tesla$
Therefore, the magnetic field at the center of solenoid is $8.73 \times {10^{ - 6}}\,Tesla$.

Now, at the end of the solenoid one angle will be zero, therefore, the magnetic field at the end of the solenoid is given by
$B = \dfrac{{4\pi \times {{10}^{ - 7}} \times \dfrac{{1000}}{{0.4}} \times \left( {5 \times {{10}^{ - 3}}} \right)}}{2}\left( {\sin 0 + \dfrac{{0.4}}{{\sqrt {{{\left( {0.4} \right)}^2} + {{\left( {0.3} \right)}^2}} }}} \right)$
$ \Rightarrow \,B = \dfrac{{4\pi \times \dfrac{{10000}}{4} \times {{10}^{ - 10}} \times 5}}{2} \times \left( {\dfrac{{0.4}}{{\sqrt {0.16 + 0.09} }}} \right)$
$ \Rightarrow \,B = \dfrac{{5\pi \times {{10}^{ - 6}}}}{2} \times \left( {\dfrac{{0.4}}{{0.5}}} \right)$
$ \Rightarrow \,B = \dfrac{{5\pi \times {{10}^{ - 6}}}}{2} \times 0.8$
$ \Rightarrow \,B = \dfrac{{4\pi \times {{10}^{ - 6}}}}{2}$
$ \Rightarrow \,B = 2\pi \times {10^{ - 6}}$
$ \Rightarrow \,B = 2 \times \dfrac{{22}}{7} \times {10^{ - 6}}\,Tesla$
$ \therefore \,B = 6.28 \times {10^{ - 6}}\,Tesla$
Therefore, the magnetic field at the end of the solenoid is $6.28 \times {10^{ - 6}}\,Tesla$.Therefore, the magnetic field on the axis at the middle and at the end of the solenoid are $8.73 \times {10^{ - 6}}\,T$ , $6.28 \times {10^{ - 6}}\,T$.

Hence, option A is the correct option.

Note:The angle at the center of the solenoid is equal because the angle on both the sides of the point will be equal. Also, we have divided the value of turns by length because the turns are on the entire length of the solenoid and we are taking the point on the solenoid. Also, the value of length and diameter is taken half in the first case because the point is at the center of solenoid.