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If the solenoid in Exercise 5.5 is free to tum about the vertical direction and a uniform hortzontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30^@ with the direction of applied field? |
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Answer» SOLUTION :Above solenoid acts like a bar magnet with magnetic dipole moment `m_(s) = 0.6 Am^(2)`. Now, torque exerted on it by uniform external magnetic FIELD is, `tau= m_(s) B sin theta` `= (0.6) (0.25) sin 30^@` `=(0.6) (0.25) (0.5)` `tau= 0.075` Nm |
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