Rankers Physics

Magnetic Properties of Matter: Practice Problem & Solution

19. An iron rod of susceptibility $599$ is subjected to a magnetising field of $1200$ A m$^{-1}$. The permeability of the material of the rod is : (2020) ($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)
$8.0 \times 10^{-5}$ T m A$^{-1}$
$2.4\pi \times 10^{-5}$ T m A$^{-1}$
$2.4\pi \times 10^{-7}$ T m A$^{-1}$
$2.4\pi \times 10^{-4}$ T m A$^{-1}$

Solution Explained:

To solve this problem, we apply the core principles of Magnetic Properties of Matter. Understanding the underlying formula is key to arriving at the correct answer below:

Relative permeability $\mu_r = 1 + \chi = 1 + 599 = 600$.
Permeability $\mu = \mu_r \mu_0 = 600 \times 4\pi \times 10^{-7} = 2400\pi \times 10^{-7} = 2.4\pi \times 10^{-4}$ T m A$^{-1}$.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. 5. A closely wound solenoid of $2000$ turns and area of cross section $1.5 \times 10^{-4} m^2$ carries a current of $2.0 A$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2} tesla$ making an angle of $30^{\circ}$ with the axis of the solenoid. The torque on the solenoid will be (2010 Mains)
2. A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the center of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the center of this coil of n turns will be: (2016 - II)
3. Assertion (A): For a perfectly diamagnetic substance, permeability is always 1. Reason (R): Small pieces of any magnetic substance are attracted by a strong magnetic field.  
4. An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the center of the circle. The radius of the circle is proportional to: (2005)
5. A long solenoid of $50text{ cm}$ length having $100$ turns carries a current of $2.5text{ A}$. The magnetic field at the centre of the solenoid is:  ($\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$) (2020)