 2 into 10 to the power minus 3 and root 3 by 0.866 and root 3 also sometime people don't even remember root 3 value because head times values are given like that okay that they will tell you delta t is 1.732 but that thing will never strike in your head okay root 3 is coming in dynamite I will cancel it one by one next question no breaks today a circular coil a u it's fun right you don't need breaks you know every time you are fighting for me circular coil of radius 10 centimeter and number of turns 500 I will give 2 minutes number of turns is 500 and okay this is radius I will call this as a because r I have to use for a distance red distance is 2 ohms okay it is placed with its plane perpendicular to the horizontal component of earth's magnetic field are you getting it you don't have to write all that don't write the story just imagine what is happening here this is only data is this much data is this much and then it is placed perpendicular to the horizontal component of earth's magnetic field so how it is placed like this horizontal component getting it it is rotated about its vertical diameter through 180 degrees it is rotated by 180 degrees in 0.25 seconds 10.25 seconds horizontal component of earth's magnetic field is 3 into 10 is 4 minus 5 tesla fine you need to get the value of emf current and the charge that has flown through the coil yeah charge that is flown through the road in this time because the emf and current how much initial flux is what initial flux is b into a into cos of what cos of 0 final flux is what b into a into cos of 180 degrees see you can choose your area vector initially like this fine once you have chosen this this area vector becomes fixed it rotates with a coil also you can't be like okay after it rotates I will chain my area vector like that no you can choose once once you have chosen like this it is fixed if coil rotates the area vector also rotates fine the final this is equal to minus ba initial is b into a so changes what minus 2 into ba fine so minus of delta 5 by delta t is the emf that is 2 ba divided by delta t this is emf e and current is what 1 by r times 2 ba by delta t and one thing you have to notice here that there are number of terms what you should do here multiply with n how you get delta q what is emf minus of d5 by dt fine current is what 1 by r d5 by dt fine current is also dq by dt right dt and dt gone so you get dq is equal to minus of d5 by r so when you integrate this you will get delta q is equal to minus of change in flux by r or magnitude of charge that has been passed across is magnitude of change in flux divided by resistance so you just divide the change in flux with the resistance you will get the charge that has flown across get the values here is 3.8 into tensor minus 3 minus 3 3.8 tensor minus 3 how much you got flux 3 pi into tensor minus 7 so 2 times of that into 500 divided by 0.25 or you made calculation that I cannot do this 3.8 into tensor minus 3 1.2 pi is 3.8 but then you are handicapped after 1.2 pi you are not able to proceed you are not able to multiply 1.2 with pi any doubt no doubt