Question

The length of a solenoid is $0.1m$ and its diameter is very small. A wire is wound over it in two layers. The number of turns in the inner layer is $50$ and that on the outer layer is $40$. The strength of current flowing in two layers in the same direction is $3$ ampere. The magnetic induction in the middle of the solenoid will be:
$(1)3.4 \times {10^{ - 3}}tesla$
$(2)3.4 \times {10^{ - 3}}gauss$
$(3)3.4 \times {10^3}tesla$
$(4)3.4 \times {10^3}gauss$

Answer 1

Hint: Electromagnetic or magnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field. The formula for magnetic induction in a solenoid is $B = \dfrac{{{\mu _o}NI}}{L}$. This can also be written as $B = {\mu _o}nI$(where $n$ is the number of turns per unit length of the solenoid).

Complete answer:
Given,
Length of the solenoid$ = 0.1m$
Number of turns in the inner layer $ = {N_1} = 50$
Number of turns in the outer layer $ = {N_2} = 40$
Current flowing in the two layers $ = 3A$
To find,
Magnetic induction in the middle of the solenoid
We will use this formula,
Magnetic induction in the middle of the solenoid $B = \dfrac{{{\mu _o}NI}}{L}$
Where,
${\mu _o} = $permeability of vacuum $ = 4\pi \times {10^{ - 7}}\dfrac{H}{m}$
$N = $ number of turns
$I = $ current flowing in the two layers
$L = $ length of the solenoid
So, the value of $B$ due to the two layers is,
$B = \dfrac{{{\mu _o}{N_1}{I_1}}}{{{L_1}}} + \dfrac{{{\mu _o}{N_2}{I_2}}}{{{L_2}}}$
Since the current and the length are the same, so the above expression becomes,
$B = \dfrac{{{\mu _o}{N_1}I}}{L} + \dfrac{{{\mu _o}{N_2}I}}{L}$
On taking $I$ and $L$ common, we get,
$B = \dfrac{{{\mu _o}I}}{L}({N_1} + {N_2})$
On putting the values which are given in the question to us, we get,
$B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 3}}{{0.1}} \times (50 + 40)$
$B = \dfrac{{4 \times 3.14 \times {{10}^{ - 7}} \times 3 \times 90}}{{0.1}}$
On further solving, we get,
$B = \dfrac{{3.14 \times {{10}^{ - 4}}}}{{{{10}^{ - 1}}}}$
$B = 3.14 \times {10^{ - 3}}T$
Thus, the magnetic induction in the middle of the solenoid will be $B = 3.14 \times {10^{ - 3}}T$
So, the final answer is $(1)3.4 \times {10^{ - 3}}tesla$.

Note: The magnetic field in a solenoid varies as the various conditions change. The magnetic field of a solenoid mainly depends upon three factors. These three factors are electric current, number of turns and the length of the solenoid which is being used.