Calculating Field Strength inside a Solenoid

Strategy

To find the field strength inside a solenoid, we use $B = μ_{0} nI$ . First, we note the number of loops per unit length is

n = \frac{N}{l} = \frac{2000}{2.00 m} = 1000 m^{- 1} = 10 {cm}^{- 1} .

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Solution

Substituting known values gives

\begin{array}{l} B & = & μ_{0} nI = (4π \times 10^{- 7} T \cdot m/A) (1000 m^{- 1}) (1600 A) \\ = & 2 .01 T. \end{array}

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Discussion

This is a large field strength that could be established over a large-diameter solenoid, such as in medical uses of magnetic resonance imaging (MRI). The very large current is an indication that the fields of this strength are not easily achieved, however. Such a large current through 1000 loops squeezed into a meter’s length would produce significant heating. Higher currents can be achieved by using superconducting wires, although this is expensive. There is an upper limit to the current, since the superconducting state is disrupted by very large magnetic fields.