Question

A long solenoid with 500 turns per unit length carries a current 1.5A.The magnetic induction at one ends of the solenoid on its axis is nearly

Answer 1

Answer:

47x10^-5

Explanation:

use the equation B = uonI divided by 2                                                                                              

                                 = 4 pi x10 ^-7 x 500 x 1.5                                                                  

                                   =4 x 3.14 x 10^-7 x 500x 1.5    ( n= 500)

          by solving we get  the above answer

Answer 2

Answer:

The magnetic induction at one end of the solenoid on its axis is nearly 4.71×10⁻⁴T.

Explanation:

The magnetic induction at the center of the solenoid is given as,

$B=\mu_0nI$           (1)

B=magnetic induction

μ₀=permitivity of free space=4π×10⁻⁷

n=number of turns per unit length

I=the current flowing in the solenoid

Whereas the magnetic induction at one of the ends is half at its center.i.e.

$B_e=\frac{B}{2}=\frac{\mu_0nI}{2}$        (2)

The values given in the question are,

n=500

I=1.5A

By substituting them in equation (2) we get;

$B_e=\frac{4\pi\times 10^-^7\times 500\times 1.5}{2}=1.5\pi \times 10^-^4=4.71\times 10^-^4T$

Hence, the magnetic induction at one end of the solenoid on its axis is nearly 4.71×10⁻⁴T.