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Two very long solenoids have the same length, but solenoid A has 18 times the number of turns, 1/5 the radius, and 7 times the current of solenoid B. Calculate the ratio of the magnetic field inside of A to that inside B.

Respuesta :

Answer:

[tex]B=126B'[/tex]

Explanation:

Let n and n' be the number of turns in solenoid A and B. r and r' be the radius of solenoid A and B. I and I' be current in solenoid A and B.

ATQ,

n = 18 n'

r = (1/5)r'

I=7I'

The magnetic field inside the solenoid is given by :

[tex]B=\mu_o nI[/tex]

So,

[tex]\dfrac{B}{B'}=\dfrac{nI}{n'I'}[/tex]

B and B' is the magnetic field in solenoid A and B.

[tex]\dfrac{B}{B'}=\dfrac{18n'\times 7I'}{n'I'}\\\\\dfrac{B}{B'}=126\\\\B=126B'[/tex]

So, the magnetic field inside the solenoid A is 126 times the magnetc field inside the solenoid B.

The ratio of magnetic field between a and b is 126.

Data;

  • L1 = L2
  • Na = 18Nb
  • Ra = 1/5Nb
  • Ia = 7Ib

Magnetic field

The magnetic field of a solenoid can be calculated using the formula

[tex]B = \frac{\mu_o N_a I_a}{L_a}[/tex]

Let's find the ratio between the field a and field b.

[tex]\frac{B_a}{B_b} = \frac{\mu_o Na I_a}{\mu_o N_b I_b} * \frac{L_b}{L_a} \\\frac{B_a}{B_b} = \frac{18N_b(7I_b}{N_bI_b}\\ \frac{B_a}{B_b} = 126[/tex]

The ratio of magnetic field between a and b is 126.

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