A long solenoid of radius 3 cm has 1100 turns per m.If the solenoid wire carries a current of 1.5 A,then calculate the magnetic field inside the solenoid. A)2.1 $\times$ 10^-3 T B)1.0 $\times$ 10^-3 T C)1.7 $\times$ 10^-4 T D)7.0 $\times$ 10^-2 T E)none of the above

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Quizplus · Accepted Answer

The magnetic field inside a long solenoid is given by the formula $B = \mu_0 n I$, where $\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}$ is the permeability of free space, $n$ is the number of turns per unit length of the solenoid, and $I$ is the current. Given $n = 1100 \, \text{turns/m} = 1100 \, \text{m}^{-1}$ and $I = 1.5 \, \text{A}$, we find $B = 4\pi \times 10^{-7} \times 1100 \times 1.5 \approx 2.07 \times 10^{-3} \, \text{T} = 2.1 \times 10^{-3} \, \text{T}$, which matches choice A.

A Long Solenoid of Radius 3 Cm Has 1100 Turns ×\times×

A Long Solenoid of Radius 3 Cm Has 1100 Turns $\times$