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ExamsJEE AdvancedPhysics

A cube of side a has identical point charges +q placed at all eight corners. The cube rotates about a symmetry axis with constant angular velocity omega. Which statement(s) about the net magnetic field at the centre of the cube is/are correct? (A) Net magnetic field at the centre is zero. (B) Net magnetic field at the centre is sqrt(2)*mu0*q*omega / (pi*a). (C) Net magnetic field at the centre is [8 / (3*sqrt(3))] * mu0*q*omega / (pi*a). (D) If the sign of any four charges is reversed (alternating), the magnetic field at the centre becomes zero.

  1. Net magnetic field at the centre is zero.
  2. Net magnetic field at the centre is sqrt(2)*mu0*q*omega / (pi*a).
  3. Net magnetic field at the centre is [8 / (3*sqrt(3))] * mu0*q*omega / (pi*a).
  4. If the sign of any four charges is reversed, the magnetic field at the centre becomes zero.

Correct answer: Net magnetic field at the centre is [8 / (3*sqrt(3))] * mu0*q*omega / (pi*a).

Solution

Each corner charge q rotates with omega; effective current I = q*omega/(2*pi). Taking the axis along a body diagonal through centre: 2 charges lie on the axis (no contribution), 6 charges orbit at perpendicular distance r_perp = a*sqrt(2)/sqrt(3)... The exact calculation for all 8 charges equally off-axis gives the net field along the rotation axis. Each charge at corner (±a/2, ±a/2, ±a/2) at distance d=a*sqrt(3)/2 from centre, each orbiting at r_perp from axis. The standard result for rotation about a face-centred axis (each corner at r_perp = a/sqrt(2), distance to centre = a*sqrt(3)/2) gives the formula in option C.

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