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Obtain an expression for Magnetic field inside a solenoid by using Ampere's Circuital Law. |
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Answer» SOLUTION :Consider a RECTANGULAR Amperian loop abcd. Along cd the field is zero because the field at the exterior is weak and moreover the field is along the axis of the SOLENOID with no PERPENDICULAR or normal component. Therefore the field outside the solenoid approaches zero. Alongtransverse sections bc and ad, the field component is zero. Let the field along be ab be B. The relevant length of the Amperian loop is l. Let n be the number of turns PER unit length. The total number of turns corresponding to the length l is nl. The enclosed current `I_(e)=Inl` where I is the current in the Solenoid.  From Ampere's circuital law `Bintdl=mu_(0)I` i.e., `Bl=mu_(0)Inl`. Therefore `B=m_(0)nI`. The direction of magnetic field is given by the right-hand rule. Note: The solenoid is used to obtain a uniform magnetic field. |
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