An iron rod of volume 10-3 m 3 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 :

Question

An iron rod of volume 10-3 m 3 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 : (A) 0.5 × 102 Am 2 (B) 50 × 102 Am 2 (C) 500 × 102 Am 2 (D) 5 × 102 Am 2

Tardigrade Admin · Accepted Answer

M =μ r NiA Here μ r = Relative permeability N = No . of turns i = Current A= Aea of cross section M =μ r Ni A =μ r n ℓ i A M =μ r niV = 1000(1000) 0.5(10-3) =500=5 × 102 Am 2

Q. An iron rod of volume $1 0^{- 3} m^{3}$ 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 :

$0.5 \times 1 0^{2} A m^{2}$

$50 \times 1 0^{2} A m^{2}$

$500 \times 1 0^{2} A m^{2}$

$5 \times 1 0^{2} A m^{2}$

$M = μ_{r} N i A$
Here
$μ_{r} =$ Relative permeability
$N = N o .$ of turns
$i =$ Current
$A =$ Aea of cross section
$M = μ_{r} N i A = μ_{r} n ℓ i A$
$M = μ_{r} niV = 1000 (1000) 0.5 (1 0^{- 3})$
$= 500 = 5 \times 1 0^{2} A m^{2}$

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

0.5×102Am2

50×102Am2

500×102Am2

5×102Am2