A long solenoid of N turns has a self-inductance L and area of cross section A. When a current i flows through the solenoid, the magnetic field inside it has magnitude B. The current i is equal to:
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Question

A long solenoid of N turns has a self-inductance L and area of cross section A. When a current i flows through the solenoid, the magnetic field inside it has magnitude B. The current i is equal to:

G1 Physics · Accepted Answer

For a solenoid, flux linkage is NΦ=N·B·A. Since L=(NΦ)/i, the current is i=B·A·N/L. Hence option A.