Question
An ideal solenoid having 5000 turns/m has an aluminium core and carries a current of 5A. If χAl = 2.3 × 10-5, then the magnetic field developed at centre will be
0.031 T
0.048 T
0.027 T
0.050 T
Solution
A.
0.031 T
Given:-
Number of turns in solenoid n = 5000 turns/m
Current carried i = 5A
Magnetic susceptibility χAl = 2.3 × 10-5
B = ?
As
B = μo ( H + A )
where,
H =
Where H = magnetic intensity
B0 = external magnetic field
μ0 = permitivity constant
=
H = ni
= 5000 × 5
= 2.5 × 104 A/m
and I = χ H
= 2.3 × 10-5 × 2.5 × 104
I = 0.575 A/m
∴ B = μo ( H + I )
= 4 × 10-7 ( 2.5 × 104 + 0.575 ) T
B = 0.031 T
