What current is required in the windings of a long solenoid that has 2470 turns

Question

What current is required in the windings of a long solenoid that has 2470 turns uni­formly distributed over a length of 0.75m in order to produce a magnetic field of magnitude 0.000201 T at the center of the solenoid? The permeability of free space is 4pi Times 10^-7T . m/A. Answer in units of mA. A magnetic field of 1 T is perpendicular to a square coil of 20 turns. The length of each side of the coil is 4 cm. Find the magnetic flux through an individ­ual turn of the coil. Answer in units of mWb. Find the magnetic flux through an individual turn of the coil if the magnetic field makes an angle of 30 degree with the normal to the plane of the coil. Answer in units of mWb.

Explanation / Answer

006)

magnetic field of solenoid

B = u0*n*I

where u0 = 4*pi*10^(-7) T.m/A

n = N/L = turns per meter

0.000201 = 4*pi*10^(-7)*(2470/0.75)*I

I = 0.000201/(4*(22/7)*10^(-7)*(2470/0.75)) = 0.0485 Amp

007)

magnetic flux = B.A = BAcos(theta)

where A is the area of the surface, and theta is the angle between the magnetic field lines and the normal (perpendicular) to A

for N turns of coil

magnetic flux = NBAcos(theta) = 20*1*(0.04*0.04)*cos(0) = 0.032 Wb = 32 mWb

008)

magnetic flux = BAcos(theta)

magnetic flux = 1*(0.04*0.04)*cos(30) = 0.00138 = 1.38 mWb

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