Question

A solenoid 83.0 cm long has a radius of 2.10 cm and a winding of 1600 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.

Answer 1

Answer:

B = 0.0087 T

Explanation:

given,                                                  

Length of solenoid (L)= 83 cm = 0.83 m            

radius of the solenoid (r)= 2.10 cm = 0.021 m

number of turns (N)= 1600 turns                    

current (I)= 3.60 A                          

Using magnetic field formula                

        $B = \dfrac{\mu_0Ni}{L}$                          

        $B = \dfrac{4 \pi \times 10^{-7} \times 1600 \times 3.6}{0.83}$

             B = 0.0087 T

the magnitude of magnetic field inside the solenoid is equal to  B = 0.0087 T