Physics · 2020
JEE Main · 4 September 2020, Shift 1 · Q14
A small bar magnet placed with its axis at 30^° with an external field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to…
A small bar magnet placed with its axis at $\displaystyle 30^{\circ}$ with an external field of $\displaystyle 0.06$ T experiences a torque of $\displaystyle 0.018$ Nm . The minimum work required to rotate it from its stable to unstable equilibrium position is :
Official answer
(2)
$\displaystyle 7.2 \times 10^{-2} \mathrm{~J}$
More from Magnetism and Matter
- Two identical small bar magnets each of dipole moment 3 √5 J / T are placed at a center to center separation of 10 cm, with their axes perpendicular…2026
- The figure gives experimentally measured B vs. H variation in a ferromagnetic material. The retentivity, co-ercivity and saturation, respectively, of…2020
- The vertical component of the earth's magnetic field is 6 × 10^-5 T at any place where the angle of dip is 37^°. The earth's resultant magnetic field…2022
- Choose the correct option:2021
- Which of the following statements are correct? Electric monopoles do not exist whereas magnetic monopoles exist. Magnetic field lines due to a…2021
- Given below are two statements: Statement I: For diamagnetic substance, -1 ≤ χ<0, where χ is the magnetic susceptibility. Statement II: Diamagnetic…2023
- A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is 1.0 T. The magnetic…2026
- A bar magnet with a magnetic moment 5.0 Am^2 is placed in parallel position relative to a magnetic field of 0.4 T. The amount of required work done…2023
JEE Main 2020 Physics question, with the answer from NTA’s final answer key. Where our answers come from.