 should I proceed with the next question type in yes or no all of you okay now let's do this question by the way the unit for magnetic field is Tesla which we know already right there is a cgs unit also cgs unit of magnetic field is goss or just g so remember this thing that one goss that is one g is 10 this power minus 4 Tesla this conversion you should remember because at times magnetic field is given in terms of goss g okay right let us proceed next question is this a short bar magnet there is a bar magnet it is placed it is placed with its axis 30 degrees with an external magnetic field this 30 degree of the bar magnet is with the magnetic field of 800 g 800 goss okay it experiences a torque of 0.016 Newton meter okay you need to find out what is the magnetic moment of the needle that is part a what is the magnetic moment of the needle second part start doing the first part okay you guys start doing this part find the magnetic moment once you get it please type in the answer in the chat box m is 0.4 I hope all of you got it this is 0.4 how you get this answer torque is mv sin theta okay so torque is given 0.016 this is equal to m into 800 g which is 800 into 10 is power minus 4 into sin theta theta is 30 degrees so sin theta is half so once you solve it you'll get the value of m to be equal to 0.4 fine this is how you do the part a in part b you need to find how much work how much work needs to be done to move this magnet from most stable position to the most unstable position do you need to do positive work or negative work you are you know increasing the energy of the system because the unstable position will have more energy right so you are going from the most stable configuration which will of course have lesser energy to the most unstable position which of course has higher energy so when you go from lower energy level to higher energy level you have to do a positive work right so there has to be a positive work now how much that should be for that whenever you have to find out work and there are energies involved always use work energy theorem which is simple you know we have done this many a times work done is equal to u2 plus k2 minus u1 plus k1 okay so I can ignore the kinetic energy I'll say that it is slowly moved and kinetic energy initially and finally are 0 this is u2 minus u1 okay and what is potential energy minus m dot b that is equal to minus of mb cos theta in a stable configuration potential energy for stable configuration is when theta is 0 so this is minus m into b and for unstable you'll have potential energy mb cos 180 degree which is minus mb into minus 1 so that is m into b so u2 is unstable so this is m into b minus of minus m into b so the answer should be 2 times m into b so when you substitute the value you'll get 0.064 this many joules of work done okay now can you find out how much current should be there in a solenoid having 1000 turns that has same magnetic dipole moment so magnetic dipole moment of a solenoid is n i a okay this should be equal to m so current is m which we have already got as 0.4 divided by number of turns which is 1000 area which is 2 into 10 raise to power minus 4 this you'll get as 2 ampere fine so I hope you are able to solve these kind of questions any doubt till now feel free to ask anything fine let us take another question you need to find the magnitude of equatorial and axial magnetic field how much they will be okay due to a bar magnet due to a bar magnet which has length 5 centimeter at a distance 50 centimeter from the center of the bar magnet okay the magnetic moment of the bar magnet is given as 0.4 ampere meter square so you need to find out what is b axial and b equatorial so b e and b a and even got the answer please message the formula for axial magnetic field is mu naught by 4 pi 2 m by r q it doesn't depend on length right we are ignoring length compared to the distance where we are finding the magnetic field where an equatorial is this correct same here so when you substitute all the values mu naught by 4 pi is 10 raise to power minus 7 this into 2 into magnetic moment which is 0.4 this divided by 0.5 into 0.5 fine so you get this as 6.4 into 10 raise to power minus 7 tesla okay equatorial will be what negative of 3.2 into 10 raise to power minus 7 tesla it is half of it fine so like this you can do this simple straightforward ones okay so this is about the bar magnet okay now we are going to discuss something which is related to the material of the magnet okay so we'll take 10 minutes break and we will meet after 10 minutes okay