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Question

A precision laboratory resistor is made of a coil of wire 1.50 cm in diameter and 4.00 cm long, and it has 500 turns. (a)What is its self-inductance? (b) What average emf is induced if the 12.0 A current through it is turned on in 5.00 ms (one-fourth of a cycle for 50 Hz AC)? (c) What is its inductance if it is shortened to half its length and counter wound (two layers of 250 turns in opposite directions)?

Answer 1

Hello Muhammad,

Self inductance of a solenoid is given as

L = µ₀ N²A/l ( L is in Henry SI unit)

where µ₀ = Permeability , N= total number of turns, A= area of cross section and l = length of solenoid

Given radius = r =0.75 cm = 0.75 * 10^-2 m, A =∏* ( 0.75*10^-2 )²

N= 500

l= 4cm = 4*10^-2 m

µ_{0 = 4∏* 10}^-7 for air

(a) Get L (in Henry) After Plugging in values above

(b) For average EMF = - L dI/dt= - L*12/(5* 10^-3 ) Plug in value of L from above to get EMF ( in Volts)

-ve sign gives only the direction of EMF

(c) Inductance will be zero if the turns are wound in opposite directions