
Find magnetic flux density at a point on the axis of a long solenoid having $5000$ turns/m when it carries a current of $2 A$.
Answer
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Hint:We have given the number of turns per unit length of the solenoid. Use the formula for magnetic field at a point of the axis of the solenoid to calculate the magnetic flux density. Express the magnetic field in tesla.
Formula used:
\[B = {\mu _0}ni\]
Here, \[{\mu _0}\] is the permeability of the free space, n is the number of turns per unit length of the solenoid and i is the current flowing through the solenoid.
Complete step by step answer:
We have the expression for magnetic flux density at a point on the axis of infinitely long solenoid,
\[B = {\mu _0}ni\]
Here, \[{\mu _0}\] is the permeability of the free space, n is the number of turns per unit length of the solenoid and i is the current flowing through the solenoid.
Substituting \[4\pi \times {10^{ - 7}}\,{\text{Wb/A}}\] for \[{\mu _0}\], 5000 turns/m for n and 2 A for i in the above equation, we get,
\[B = \left( {4\pi \times {{10}^{ - 7}}} \right)\left( {5000} \right)\left( 2 \right)\]
\[ \therefore B = 0.0126\,{\text{T}}\]
Therefore, the magnetic flux density at a point on the axis of the long solenoid is 0.0126 T.
Additional information:
The magnetic field along the axis of an ideal solenoid is uniform while the magnetic field at some distance away from the axis of solenoid becomes zero. We can determine the magnetic flux density along the axis of the solenoid by applying Ampere’s law, \[\oint {\vec B \cdot d\vec l} = {\mu _0}I\], where, I is the net current encircled by the Amperian loop.
Note:The expression, \[B = {\mu _0}ni\] is valid for ideal solenoid which has infinite length, however, it can also be applied for realistic solenoids having finite length. The magnetic flux density remains the same along the axis of the solenoid and it weakens as we move away from the central axis outside the solenoid. While solving these types of questions, it is crucial to note that, in the above expression, n is the number of turns per unit length of the solenoid and not just the number of turns.
Formula used:
\[B = {\mu _0}ni\]
Here, \[{\mu _0}\] is the permeability of the free space, n is the number of turns per unit length of the solenoid and i is the current flowing through the solenoid.
Complete step by step answer:
We have the expression for magnetic flux density at a point on the axis of infinitely long solenoid,
\[B = {\mu _0}ni\]
Here, \[{\mu _0}\] is the permeability of the free space, n is the number of turns per unit length of the solenoid and i is the current flowing through the solenoid.
Substituting \[4\pi \times {10^{ - 7}}\,{\text{Wb/A}}\] for \[{\mu _0}\], 5000 turns/m for n and 2 A for i in the above equation, we get,
\[B = \left( {4\pi \times {{10}^{ - 7}}} \right)\left( {5000} \right)\left( 2 \right)\]
\[ \therefore B = 0.0126\,{\text{T}}\]
Therefore, the magnetic flux density at a point on the axis of the long solenoid is 0.0126 T.
Additional information:
The magnetic field along the axis of an ideal solenoid is uniform while the magnetic field at some distance away from the axis of solenoid becomes zero. We can determine the magnetic flux density along the axis of the solenoid by applying Ampere’s law, \[\oint {\vec B \cdot d\vec l} = {\mu _0}I\], where, I is the net current encircled by the Amperian loop.
Note:The expression, \[B = {\mu _0}ni\] is valid for ideal solenoid which has infinite length, however, it can also be applied for realistic solenoids having finite length. The magnetic flux density remains the same along the axis of the solenoid and it weakens as we move away from the central axis outside the solenoid. While solving these types of questions, it is crucial to note that, in the above expression, n is the number of turns per unit length of the solenoid and not just the number of turns.
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