
A solenoid of length $1.5m$ and $4cm$ diameter possesses $10\text{ turns/cm}$. A current of $5A$ is flowing through it, the magnetic induction at the axis of the solenoid is \[\left( {{\mu }_{0}}=4\pi \times {{10}^{-7}}weber.am{{p}^{-1}}{{m}^{-1}} \right)\]
$A)\text{ }4\pi \times {{10}^{-5}}gauss$
$B)\text{ 2}\pi \times {{10}^{-5}}gauss$
$C)\text{ }4\pi \times {{10}^{-5}}tesla$
$D)\text{ 2}\pi \times {{10}^{-3}}tesla$
Answer
594.9k+ views
Hint: This problem can be solved by using the direct formula for the magnitude of the magnetic induction inside a solenoid at the axis in terms of the number of turns per unit length and the current through the solenoid. We will plug in the information given in the question in the formula to get the required answer.
Formula used:
$B={{\mu }_{0}}nI$
Complete step by step answer:
Let us use the direct formula for the magnetic field inside a solenoid at its axis.
The magnitude of magnetic induction $B$ inside a solenoid which carries a current $I$ and has $n$ number of turns per unit length is given by
$B={{\mu }_{0}}nI$ --(1)
where \[\left( {{\mu }_{0}}=4\pi \times {{10}^{-7}}weber.am{{p}^{-1}}{{m}^{-1}} \right)\]is the permeability of free space.
Now, let us analyze the question.
The current carried by the solenoid is $I=5A$.
The number of turns per unit length of the solenoid is $n=10\text{ turns/cm = }10\times {{10}^{2}}\text{ turns/m = }{{10}^{3}}\text{ turns/m}$ $\left( \because 1c{{m}^{-1}}={{10}^{2}}{{m}^{-1}} \right)$
Let the magnitude of the magnetic induction at the axis of the solenoid be $B$.
Therefore, using (1), we get
$B={{\mu }_{0}}nI=4\pi \times {{10}^{-7}}\times {{10}^{3}}\times 5=2\pi \times {{10}^{-3}}T$
Hence, we have got the required value for the magnitude of the magnetic induction at the axis of the solenoid.
So, the correct answer is “Option D”.
Note:
Students must not get confused upon seeing the values for the length and the diameter of the solenoid given and must remember that the formula for the magnetic induction of a solenoid at the axis does not require them. Only in the case the total number of turns is given, it can be divided by the total length to get the number of turns per unit length. Sometimes extra information is given in questions to put doubts in the minds of students who do not know the concept clearly and lead them into making a mistake.
Formula used:
$B={{\mu }_{0}}nI$
Complete step by step answer:
Let us use the direct formula for the magnetic field inside a solenoid at its axis.
The magnitude of magnetic induction $B$ inside a solenoid which carries a current $I$ and has $n$ number of turns per unit length is given by
$B={{\mu }_{0}}nI$ --(1)
where \[\left( {{\mu }_{0}}=4\pi \times {{10}^{-7}}weber.am{{p}^{-1}}{{m}^{-1}} \right)\]is the permeability of free space.
Now, let us analyze the question.
The current carried by the solenoid is $I=5A$.
The number of turns per unit length of the solenoid is $n=10\text{ turns/cm = }10\times {{10}^{2}}\text{ turns/m = }{{10}^{3}}\text{ turns/m}$ $\left( \because 1c{{m}^{-1}}={{10}^{2}}{{m}^{-1}} \right)$
Let the magnitude of the magnetic induction at the axis of the solenoid be $B$.
Therefore, using (1), we get
$B={{\mu }_{0}}nI=4\pi \times {{10}^{-7}}\times {{10}^{3}}\times 5=2\pi \times {{10}^{-3}}T$
Hence, we have got the required value for the magnitude of the magnetic induction at the axis of the solenoid.
So, the correct answer is “Option D”.
Note:
Students must not get confused upon seeing the values for the length and the diameter of the solenoid given and must remember that the formula for the magnetic induction of a solenoid at the axis does not require them. Only in the case the total number of turns is given, it can be divided by the total length to get the number of turns per unit length. Sometimes extra information is given in questions to put doubts in the minds of students who do not know the concept clearly and lead them into making a mistake.
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