Solenoid 2 has twice the diameter, twice the length, and twice as many turns as solenoid 1. How does the field B2 at the center of solenoid 2 compare to B1 at the center of solenoid 1?

Respuesta :

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  option 3

Explanation:

From the question we are told that

   The diameter of solenoid 1 is  [tex]d_1[/tex]

   The length of solenoid 1 is   [tex] L_1 [/tex]

    The  number of turns of solenoid is  [tex]N_1 [/tex]

   The diameter of solenoid 2 is  [tex]d_2 = 2d_1[/tex]

   The length of solenoid 2 is   [tex] L_2 = 2L_1 [/tex]

    The  number of turns of solenoid  2 is    [tex] N_2 = 2 N_1 [/tex]

Generally the magnetic in a solenoid is mathematically represented as

     [tex]B  =  \frac{\mu_o *  N  *  I }{L}[/tex]

From this equation we see that

     [tex]B  \ \alpha \  \frac{N}{L}[/tex]

     [tex]B   =  C   \frac{N}{L}[/tex]

Here C stands for constant

=>   [tex]C =  \frac{B *  \frac{L}{N}[/tex]

=>    [tex]\frac{B_1 *  \frac{L_1}{N_1}   = \frac{B_2 *  \frac{L_2}{N_2}   [/tex]

=>  [tex]\frac{B_1}{B_2 }  =  \frac{N_1 L _2}{ N_2L_1}[/tex]

=>   [tex]\frac{B_1}{B_2 }  =  \frac{N_1 * (2 L_1)}{ (2 N_2)L_1}[/tex]

=>   [tex]\frac{B_1}{B_2 }  =  1[/tex]

=>   [tex]B_2 = B_1[/tex]

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