Help A long Solenoid with n = 900.0 turns/m has a radius of r= 10.0 cm and carries a current of I=0.5 A. Thissolenoid has a length of 7 = 1.2 m and the wiring resistance is equivalent to R = 2.49a) What is the magnetic field produced inside this solenoid?b) If we bend this solenoid into a toroid, what is the magnetic field inside the toroid?wwwNAWorking with the solenoid, we next turn off the current source to it and immerse it in a uniform magnetic filedparallel to the axis of the solenoid, with its value changing with a frequency of f-60 Hz, by B = B. Cosat.d) What is the heat power generated in this solenoid circuit?c) Calculate the flux of this B-filed through the solenoid as a function of time and use the Faraday's law todetermine the value of Bo such that the maximum generated current is 0.5 A in the solenoid. (At this time,the two ends of the solenoid wire is connected to form a closed circuit) e) Going back to part (a), we make the steady current to suddenly reduce by a factor of two in 5 us. what isthe direction and magnitude of the electric field at the distance of r= 5.0 cm away from the axis ofsolenoid?Induced Eside viewelleChanging Solenoid currentFaraday's law describes an induced electric field.looking into solenoidChanging B inside solenoidf) In this case, what is the direction and magnitude of the electric field at the distance of 25.0 cm away fromthe axis of solenoid?

Note: As per company policy only first 3 are solved below.Given:turn density of solenoid, n = 900 turns/mradius , r = 10.0 cm = 0.1 mcurrent in solenoid, I = 0.5 Alength of solenoid, l = 1.2 mwire resistance, R = 2.4 ΩAsked:a) Magnetic field inside the solenoid, B = ?b) magnetic field in the toroid, Bt = ?c) flux through the solenoid, Φ =?Solution:a)Magnetic field in a solenoid is given byB = μo×n×IB = 4π×10−7×900×0.5B = 0.5655 mTb)Magnetic field inside a toroid is given asBt = μoNI2πr'here N is total number of turnsand r' is the radius of toroidwe know that circumference of toroid is 2πr'and since the solenoid of l = 1.2 m is made into a toroid 2πr' = l = 1.2 mhenceBt = μoNl×IBt = μo×n×I hence Bt = B Bt = 0.5655 mTc)Now flux is given asΦt = N×Bt×Ahere N is total turns on the solenoidN = n×l = 900×1.2 = 1080 turnsBt is magnetic field, Bt = Bocosωthere frequency of B is f = 60 Hzhence ω = 120π rad/sthereforeBt = Bocos120πtand A is area A = πr2 = π×0.12 = 0.01π m2thereforeΦt = 1080×Bocos120πt×0.01π Φt = 33.929×Bocos120πtNow to find Bo we will use faradays lawE = −dΦtdt = −33.929×Bo×120π×−sin120πtE = 12791.01×Bosin120πtit is said that a max current of Io = 0.5 Anow from ohms lawE = It×R = 12791.01×Bosin120πthenceIo×R = 12791.01×BoBo = 0.5×2.412791.01Bo = 93.816×10−6 Thence the flux is Φt =33.929×Bocos120πt Φt = 33.929×93.816×10−6×cos120πtΦ(t) = 3.183cos(120πt) mWb

Answered: A long Solenoid with n=900.0 turns/m…

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A long Solenoid with n=900.0 turns/m has a radius of r= 10.0 cm and carries a current of I=0.5 A. This solenoid has a length of 1 = 1.2 m and the wiring resistance is equivalent to R = 2.492