A Solenoid of length 20cm and radius 0.8cm with 400 turns is in an external magn

Question

A Solenoid of length 20cm and radius 0.8cm with 400 turns is in an external magnetic field of 575G that makes and angle of 50degrees with the axis of the solenoid. what is the flux through the solenoid?

b) For the circiut above, B=0.05T, R=0.1 Ohms, I= 0.3A and L= 10 cm at time t=0s. What is the velovity of the bar at the time=0 s?

please show me how to do this

A Solenoid of length 20cm and radius 0.8cm with 400 turns is in an external magnetic field of 575G that makes and angle of 50degrees with the axis of the solenoid. what is the flux through the solenoid? b) For the circiut above, B=0.05T, R=0.1 Ohms, I= 0.3A and L= 10 cm at time t=0s. What is the velovity of the bar at the time=0 s? please show me how to do this

Explanation / Answer

Area

A=pi*(0.8*10-2)2=2.01*10-4 m2

Magnetic flux

phi=NBACos(o) =400*(575*10-4)*(2.01*10-4)*cos50

phi=2.97*10-3 wb or 2.97 mwb

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