Courses
Courses for Kids
Free study material
Offline Centres
More
Store

A long solenoid has $200$ turns per $cm$ and carries a current $i$ . The magnetic field at its centre is $6.28 \times {10^{ - 2}}Wb{m^{ - 2}}$ . Another long solenoid has $100$ turns per $cm$ and it carries a current $i/3$ . The value of magnetic field at its centre isA. $1.05 \times {10^{ - 2}}Wb{m^{ - 2}}$ B. $1.05 \times {10^{ - 5}}Wb{m^{ - 2}}$ C. $1.05 \times {10^{ - 3}}Wb{m^{ - 2}}$ D. $1.05 \times {10^{ - 4}}Wb{m^{ - 2}}$

Last updated date: 13th Jun 2024
Total views: 384.3k
Views today: 6.84k
Verified
384.3k+ views
Hint: Using the Ampere’s circuital law establishes the relation for the magnetic field inside a solenoid. From that equation we can calculate the current flowing in the first solenoid. Then substitute this value of current for the second solenoid.

Formulas used:
$\oint {\overrightarrow B } \cdot d\overrightarrow l = {\mu _0}i$ where $B$ is the magnetic field around any closed path, ${\mu _0}$ is the permeability of free space and $i$ is the current flowing through the area enclosed by the path.
$B = {\mu _0}ni$ where $n$ is the number of turns of the coil per unit length.

A solenoid is a type of electromagnet made by wounding coils into a tightly packed helix. When current flows through the conducting coil, a magnetic field is generated inside the solenoid such that it behaves as a magnet. We can calculate the strength of the magnetic field inside the solenoid using Ampere’s law.
Ampere’s circuital law states that the line integral of the magnetic field $\overrightarrow B$ around any closed oath is equal to ${\mu _0}$ times the net current $i$ flowing through the area enclosed by the path. That is,
$\oint {\overrightarrow B } \cdot d\overrightarrow l = {\mu _0}i$
where ${\mu _0}$ is the permeability of free space (constant).
Using this law we can establish the relation of the magnetic field at the centre of a long solenoid which is given by the expression $B = {\mu _0}ni$ where $n$ is the number of turns of the coil per unit length.
According to the question we have, $B = 6.28 \times {10^{ - 2}}Wb{m^{ - 2}}$ and $n = 200 \times {10^2}{m^{ - 1}}$
$B = {\mu _0}ni$
$\Rightarrow$ $i = \dfrac{B}{{{\mu _0}n}}$
$\Rightarrow i = \dfrac{{6.28 \times {{10}^{ - 2}}}}{{{\mu _0} \times 200 \times {{10}^2}}}$
Now for the second solenoid, we have $n = 100 \times {10^2}{m^{ - 1}}$
Thus, substituting the value of $i$ from the previous equation we have,
${B_2} = {\mu _0} \times 100 \times {10^2} \times \dfrac{{6.28 \times {{10}^{ - 2}}}}{{{\mu _0} \times 200 \times {{10}^2} \times 3}}$
$\Rightarrow {B_2} = 1.05 \times {10^{ - 2}}Wb{m^{ - 2}}$
Therefore, the correct option is A.

Note
The field $\overrightarrow B$ is independent of the length and the diameter of the solenoid and is uniform over the cross-section of the solenoid. The uniform magnetic field within a long solenoid is parallel to the solenoid axis. Also the expression used in this solution to calculate the magnetic field is only valid for a very long solenoid. For a solenoid with definite length, the formula is different.