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A closely wound solenoid of 2000 turns and are of cross - section 1.6xx10^(-4)m^2 carrying a current of 4.0 A , is suspended through its centre allowing it to turn in a horizontal plane. a. What is the magnetic moment associated with the solenoid ? b. What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5xx10^(-2)T is set up at an angle of 30^@ with the axis of the solenoid ? |
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Answer» SOLUTION :N = 200 , A`1.6xx10^(-4)m^2,I=4.0A` a. `m=ANI = 1.6 xx10^(-4) xx2000xx4.0=1.28Am^2` , along the axis b. `B = 7.5 xx10^(-2)T, theta = 30^@` NET force = 0 `tau=mBsintheta=1.28xx7.5xx10^(-2)xxsin30=0.64xx7.5xx10^(-2)=4.800xx10^(-2)Nm` By the ACTION of this, the solenoid can come to the direction of external FIELD. |
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