Question

A solenoid of length 50 cm has 5000 turns/m and carries a current of 5 A. What
is the magnitude of the magnetic field inside the solenoid ?
​

Answer 1

Explanation:

Here,

I=2.5A,l=50cm=0.50 m andn=

0.50

100

=200m

−1

∴ B=

2

μ

0

nI

=

2

4π×10

−7

×200×2.5

= 3.14 x 10

4

T