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Question #43289

Difficulty: 8/10 | 🟢 Free

The Question

A long cylindrical bar magnet of mass $m_0$ , length $L$ , and uniform magnetization falls vertically along the axis of a long, thin-walled copper tube of radius $R$ (where $L \gg R$ ). Due to electromagnetic damping, it reaches a constant terminal velocity $v_0$ . The magnet is then cut transversely (perpendicular to its length) into two identical halves. One of these halves is dropped into the same copper tube. Assuming the length of the half-magnet is still much greater than the tube's radius ( $L/2 \gg R$ ), what is its new terminal velocity $v_1$ ? Assume the tube is infinitely long and air resistance is negligible.

Options

A.
$v_1 = \frac{v_0}{2}$
B.
$v_1 = 2 v_0$
C.
$v_1 = v_0$
D.
$v_1 = \frac{v_0}{4}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Effective Moment of Cut Magnets|Faraday’s Law Vector Form|Induced Electric Fields

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of magnetic flux linkage for extended dipoles versus point dipoles.

A common trap is to assume the drag force is proportional to the square of the magnetic moment ( $F \propto M^2 v$ ), which is only true for a point dipole ( $L \ll R$ ).

If a student incorrectly applies the point dipole model, they will deduce that halving the mass and halving the magnetic moment leads to $v_1 \propto m/M^2 \propto (m_0/2)/(M_0/2)^2 = 2 v_0$ .

However, for a long magnet ( $L \gg R$ ), the flux linkage is dominated by the independent magnetic poles.

Since a transverse cut leaves the pole strength $q_m$ unchanged, the drag coefficient remains identical, and the terminal velocity simply scales with the reduced mass, yielding $v_1 = v_0 / 2$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Magnetism and Matter - Class 12
Concept	Effective Moment of Cut Magnets
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free