Exams › JEE Main › Physics

An iron rod of volume 10⁻³ m³ and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be -

1. 0.5 × 10² A m²
2. 50 × 10² A m²
3. 500 × 10² A m²
4. 5 × 10² A m²

Correct answer: 5 × 10² A m²

Solution

The magnetic moment of the rod is calculated using the formula M = nIA, where n is the number of turns per unit length, I is the current, and A is the cross-sectional area of the rod. Given the parameters, the calculations yield a magnetic moment of 5 × 10² A m², confirming option D as the correct answer.

Related JEE Main Physics questions

• The SI unit Weber is used for measuring which of the following quantities?
• Two identical magnetic dipoles, each having magnetic moment 1.0 A·m², are kept 2 m apart with their axes mutually perpendicular. The magnetic field at the midpoint between them is
• A material has relative permittivity εr and relative permeability μr. For a diamagnetic substance, which pair of values is physically permissible?
• A bar magnet freely suspended in the horizontal component of Earth’s magnetic field H has a time period of 4 s. If a second magnet is placed nearby and the time period becomes 2 s, then the magnetic field produced by the second magnet is:
• Which magnetic properties should the core material possess to be suitable for an electromagnet?
• A bar magnet has a length much greater than its width and thickness. In a vibration magnetometer, its oscillation period is 2 s. The magnet is then divided lengthwise into three equal pieces, and the three pieces are stacked together so that like poles are in contact. What is the oscillation period of this arrangement?

⚔️ Practice JEE Main Physics free + battle 1v1 →