InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A rectangular coil of n turns each of area A, carrying current I, when suspended in a uniform magnetic field B, experiences a torque `tau=nIBAsin theta` where `theta` is the angle with a normal drawn on the plane of coil makes with the direction of magnetic field. This torque tends to rotate the coil and bring it in equilibrium position. In stable equilibrium state, the resultant force on the coil is zero. The torque on coil is also zero and the coil has minimum potential energy. Read the above passage and answer the following questions: (i) In which position, a current carrying coil suspended in a uniform magnetic field experiences (a) minimum torque and (b) maximum torque? (ii) A circular coil of 200 turns, radius `5cm` carries a current of `2*0A`. It is suspended vertically in a uniform horizontal magnetic field of `0*20T`, with the plane of the coil making an angle of `60^@` with the field lines. Calculate the magnitude of the torque that must be applied on it to prevent it from turning. (iii) What is the basic value displayed by the above study? |
|
Answer» (i) As `tau=nIBA sin theta`, therefore, (i) `tau=0`, when `sin theta=0` or `theta=0^@`, i.e., when the plane of coil is perpendicular to the direction of magnetic field. (ii) `tau=`maximum, when `sin theta`=maximum `=1` or `theta=90^@` `tau_(max)=nIBAxx1=nIBA` It will be so when the plane of coil is parallel to the direction of magnetic field. (ii) Here, `n=200, r=0*05m, I=2*0A, B=0*20T, theta=90^@-60^@=30^@`. `tau=nIBA sin theta=nIB(pir^2)sin theta=200xx2*0xx0*20[(22//7)xx(0*05)^2]xxsin 30^@` `=0*314N-m=0*31Nm`. (iii) From the above study, we find that when potential energy of the coil is minimum, both force and torque acting on the coil are zero. The same is ture in real life. They say "Nanak neevan jo chale, lage na tatti va", i.e., a person who is humble and boats of nothing, would be a happy person, with no pulls and pressures of life. |
|