10.50 cm long solenoid has a radius of 1 c.m and is made up of 500 turns .If it carries acurrent of 5Ampere then ,Find the magnitude of magnetic field inside the solenoid.

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Solution

Dear Student,
$N u m b e r o f t u r n s p e r u n i t l e n g t h, n = \frac{N}{l e n g t h o f t h e s o l e n o i d} = \frac{500}{10.50 \times 10^{- 2}} = 4762 t u r n s / m M a g n e t i c f i e l d i n s i d e t h e s o l e n o i d, B = μ_{o} n I w h e r e, μ_{o} = 4 π \times 10^{- 7} T e s l a / A m I = 5 A S o, B = (4 π \times 10^{- 7} T e s l a / A m) \times (4762 t u r n s / m) \times 5 A = 5.98 \times 10^{- 3} T$

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