Question

The magnetic field at a point inside a long current-carrying solenoid compared to the magnetic field at the axial end point is:

• A. half
• B. twice
• C. same
• D. four times

Hint:

Magnetic field inside a long solenoid: \(B = \mu_0 n I\). At the end, it drops to half: \(B_\text{end} \approx \frac{1}{2} B_\text{inside}\). Use symmetry and superposition to estimate end fields.

Answer 1

The Correct Option is B

Step 1: Field inside a long solenoid.
For an ideal solenoid, B_inside = μ₀ n I, uniform and axial.

Step 2: Field at the axial end.
At the solenoid's end face, the field drops to roughly half: B_end ≈ (1/2) μ₀ n I, derived from summing loop contributions.

Step 3: Ratio comparison.
B_inside / B_end = (μ₀ n I) / (½ μ₀ n I) = 2.

Step 4: Physical reasoning.
Inside, all turns add constructively; at the end, only about half the turns reinforce axially.

Step 5: Conclusion.
The interior field is twice the field at the axial end.

Step 6: Additional remark.
In finite solenoids, the field tapers smoothly from the uniform central region to half-value at the ends, consistent with Ampère's law.