wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A solenoid 30cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is 1.5cm2. If the relative permeability of the iron is 6000. What is the self inductance of the solenoid?


A

15H

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

2.5H

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

3.5H

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

0.5H

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

The correct option is A

15H


Step 1: Given data

Length of a solenoid, L=30cm=0.3m

Cross sectional area of the solenoid, A=1.5cm2=1.5×10-4m2

Number of loops = 2000

Relative permeability of the iron μr=6000

Let permeability of free space μ0=4π×10-7H/m

Step 2: Find self inductance of the solenoid

Permeability is the measure of magnetization that a material obtains in response to an applied magnetic field. A relative permeability is defined as the ratio of the material permeability to the permeability of free space (or vacuum)

μr=μμ0

Where,

μ=permeability of substance

μ=μ0×μrμ=4π×10-7×6000μ=75.42×10-4H/m

Self-inductance of a solenoid is defined as the flux associated with the solenoid when a unit amount of current is passed through it. It is given by

=μ0μrN2AL=4π×10-7×6000×20002×1.5×10-40.3=15H

Hence, option A is correct.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Force on a Current Carrying Conductor in a Magnetic Field
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon