Two ideal solenoids of radii R and 4 R , respectively, have n turns per meter, and each carries a current of I . Both currents flow in the same direction. The small‑radius solenoid is placed inside the large-radius solenoid so that their axes of symmetry are parallel and separated by a distance d = 2 R . Neglecting any magnetic screening effects, express the magnetic field strength on the axis of the small-radius solenoid analytically in terms of the quantities given and the magnetic permeability of a vacuum, μ 0 .
The Correct Answer and Explanation is :
To determine the magnetic field strength on the axis of the small-radius solenoid due to both solenoids, we can use the formula for the magnetic field inside a solenoid and apply the principle of superposition.
Magnetic Field of a Solenoid
The magnetic field ( B ) inside an ideal solenoid is given by:
[
B = \mu_0 n I
]
where ( \mu_0 ) is the permeability of free space, ( n ) is the number of turns per unit length, and ( I ) is the current flowing through the solenoid.
Magnetic Field Due to the Small Solenoid
For the small solenoid (radius ( R ) and length ( L )), the magnetic field at its center (on its axis) is:
[
B_{\text{small}} = \mu_0 n I
]
Magnetic Field Due to the Large Solenoid
For the large solenoid (radius ( 4R ) and the same ( n )), the magnetic field at its center is also given by the same formula, since the magnetic field inside a long solenoid is uniform. Therefore, at the axis of the small solenoid (which is offset by ( d = 2R )), the magnetic field ( B_{\text{large}} ) can be determined by recognizing that the magnetic field inside the large solenoid is uniform and does not diminish significantly due to the distance, provided ( R ) is much smaller than the length of the solenoid:
[
B_{\text{large}} = \mu_0 n I
]
Total Magnetic Field on the Axis of the Small Solenoid
Now, since both currents flow in the same direction, we can simply add the contributions from each solenoid at the location of the small solenoid. Therefore, the total magnetic field ( B_{\text{total}} ) at the axis of the small solenoid is:
[
B_{\text{total}} = B_{\text{small}} + B_{\text{large}} = \mu_0 n I + \mu_0 n I = 2\mu_0 n I
]
Conclusion
Thus, the magnetic field strength on the axis of the small-radius solenoid can be expressed as:
[
B_{\text{total}} = 2 \mu_0 n I
]
This result reflects the cumulative effect of both solenoids’ magnetic fields on the axis of the smaller one, demonstrating how the superposition principle applies in magnetic field interactions. Each solenoid contributes equally to the total magnetic field due to their identical turn density and current direction, resulting in a doubled effect at the center of the smaller solenoid.