Answer:
The current the solenoid carries is as large as 1591.3 A
Explanation:
Given;
length of solenoid, L = 2.0m
radius of the solenoid, R = 30 cm = 0.3 m
number of turns of wire, N = 1000
strength of magnetic field, B = 1.0 T
The magnetic field strength inside a solenoid is;
[tex]B = \mu_o nI\\\\[/tex]
Where;
μ₀ is permeability of free space, = 4π x 10⁻⁷
n is number of turns per length
I is the current in the solenoid
[tex]B = \mu_o nI\\\\B = \mu_o \frac{N}{L} I\\\\B = \frac{\mu_oNI}{L} \\\\I = \frac{BL}{\mu_oN} \\\\I = \frac{1*2 }{4\pi* 10^{-7}*1000} \\\\I = 1591.3 \ A[/tex]
Therefore, the current the solenoid carries is as large as 1591.3 A
