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: Recall magnetic field inside a solenoid.
For an ideal long solenoid carrying current \(I\) with \(n\) turns per unit length, the magnetic field at the center is: \[ B_\text{inside} = \mu_0 n I \] This is uniform and directed along the axis of the solenoid.

Step 2: Magnetic field at the end of solenoid.
At the axial end of the solenoid, the magnetic field is approximately half of that at the center: \[ B_\text{end} \approx \frac{1}{2} \mu_0 n I \] This comes from integrating contributions of each circular loop along the solenoid’s length.

Step 3: Compare fields.
\[ \frac{B_\text{inside}}{B_\text{end}} = \frac{\mu_0 n I}{\frac{1}{2} \mu_0 n I} = 2 \]

Step 4: Physical reasoning.
Inside a long solenoid, the fields from all loops add constructively along the axis. At the end, only half of the loop contributions reinforce in the axial direction, halving the field.

Step 5: Conclusion.
Thus, the magnetic field at a point inside the solenoid is twice the field at the axial end.

Step 6: Practical note.
For finite solenoids, the field inside is nearly uniform except near the ends where it decreases smoothly to half its central value. This is consistent with Ampère's law and superposition principle.