Answer:
6
Explanation:
The magnetic field inside a solenoid is given by the following formula:
[tex]B = \mu_{0}nI[/tex]
where,
B = Magnetic Field Inside Solenoid
μ₀ = permittivity of free space
n = No. of turns per unit length
I = Current Passing through Solenoid
For Solenoid 1:
[tex]B_{1} = \mu_{0}n_{1}I ------------------- equation 1[/tex]
For Solenoid 2:
n₂ = 6n₁
Therefore,
[tex]B_{1} = \mu_{0}n_{2}I\\B_{1} = 6\mu_{0}n_{1}I ----------------- equation 2[/tex]
Diving equation 1 and equation 2:
[tex]\frac{B_{2}}{B_{1}} = \frac{6\mu_{0}nI}{\mu_{0}nI}\\\\\frac{B_{2}}{B_{1}} = 6[/tex]
Hence, the correct option is:
6
The ratio of the magnetic fields in interior 2 to interior 1 will be
[tex]\dfrac{B_2}{B_1} =\dfrac{6}{1}[/tex]
The formula for the magnetic fields inside the solenoid will be given:
[tex]B=\mu_onI[/tex]
here,
B = Magnetic Field Inside Solenoid
μ₀ = permittivity of free space
n = No. of turns per unit length
I = Current Passing through Solenoid
For the first Solenoid
[tex]B_1=\mu_on_1I[/tex]..................(1)
For the second solenoid
[tex]n_2=6n_1[/tex]
Now
[tex]B_2=\mu_on_2I[/tex]
[tex]B_2=\mu_o(6n_1)I[/tex]..................(2)
Diving equation 1 and equation 2:
[tex]\dfrac{B_2}{B_1} =\dfrac{\mu_o(6n_1)I}{\mu_on_2I}[/tex]
[tex]\dfrac{B_2}{B_1} =6[/tex]
Thus the ratio of the magnetic fields in interior 2 to interior 1 will be
[tex]\dfrac{B_2}{B_1} =\dfrac{6}{1}[/tex]
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