AP EAMCET20227 Jul 2022Morning ShiftPhysicsMagnetic Properties of MatterActual
A short bar magnet produces a magnetic field of 6.4 10⁻⁵ ~T at a distance of 20 ~cm from the centre of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of 40 ~cm from the centre of the magnet on the axis, is
Options
- A4.8 10⁻⁵ ~T
- B3.2 10⁻⁵ ~T
- C1.6 10⁻⁵ ~T
- D6.4 10⁻⁵ ~T
Correct answer
C. 1.6 10⁻⁵ ~T
Step-by-step solution
Magnetic field produced by a magnet on its axis and its equator are given by B_ axis = ₀ 4 2 M r₁^3 B_ equator = ₀ 4 = M r₂^3 B_ axis =B_ equator 2 r₂^3 r₁^3 ...(i) Where, M = magnetic dipole moment of magnet and r= distance from centre of magnet So, B_ axis B_ equator = ₀ 4 2 M r₁^3 ₀ 4 M r₂^3 = 2 r₂^3 r₁^3 B_ axis =B_ equator 2 r₂^3 r₁^3 ...(ii) Here, given B_ equator =6.4 10⁻⁵ ~T aligned & r₁=40 ~cm =40 10⁻² ~m & r₂=20 ~cm =20 10⁻² ~m aligned So from eq. (i), we have, B_ axis = 6.4 10⁻⁵ 2 (20 10⁻² )^3 (40 10⁻² )