A magnet has a dimensions of 25 cm - 10 cm - 5 cm and pole strength of 200- 10-3Am The intensity of magnetization due to it is -

The correct option is D 40A/m P=200× 10^ 3 Am l=25 cm A=10× 5× 10^ 4 m^2 I=P× lA× l =200× 10^ 310× 5× 10^ 4 =40 A/m ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Earth's Magnetism

A magnet has ...

Question

A magnet has a dimensions of 25 cm $\times$ 10 cm $\times$ 5 cm and pole strength of $200 \times 10^{- 3}$ Am The intensity of magnetization due to it is :

6.25A/m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

62.5A/m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40A/m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4A/m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 40A/m
$P = 200 \times 10^{- 3} A m$
$l = 25 c m$
$A = 10 \times 5 \times 10^{- 4} m^{2}$
$I = \frac{P \times l}{A \times l}$
$= \frac{200 \times 10^{- 3}}{10 \times 5 \times 10^{- 4}}$
$= 40 A / m$

Suggest Corrections

Similar questions

The dimensions of a bar magnet are $10 c m \times 20 c m \times 20 c m$ . It's pole strength is $200$ $A m$ . It's intensity of magnetization is :

Q. The

10 c m

long magnet has pole strength of

12 A m

. The magnetic field at a distance

20 c m

from its centre :

Q. The magnetic moment of a magnet of dimensions

(10 cm \times 4 cm \times 1 cm)

4 {A-m}^{2}

. Its intensity of magnetisation is :

Q. The magnetic moment of a magnet of length 10 cm and pole strength 4.0 Am will be

A long bar magnet has a pole strength of 10 Am. Find the magnetic field at a point on the axis of the magnet distance of 5 cm from the north pole of the magnet.

Join BYJU'S Learning Program

A magnet has a dimensions of 25 cm × 10 cm × 5 cm and pole strength of 200×10−3Am The intensity of magnetization due to it is :

The correct option is D 40A/mP=200×10−3Aml=25cmA=10×5×10−4m2I=P×lA×l=200×10−310×5×10−4=40A/m

A magnet has a dimensions of 25 cm $\times$ 10 cm $\times$ 5 cm and pole strength of $200 \times 10^{- 3}$ Am The intensity of magnetization due to it is :

The correct option is D 40A/m
$P = 200 \times 10^{- 3} A m$
$l = 25 c m$
$A = 10 \times 5 \times 10^{- 4} m^{2}$
$I = \frac{P \times l}{A \times l}$
$= \frac{200 \times 10^{- 3}}{10 \times 5 \times 10^{- 4}}$
$= 40 A / m$