Question

Physics
13) A wall watch [suspended on the wall from east to west] the length
of its seconds arm is 14cms. Calculate the potential difference
created between the ends of the arm if the horizontal component
of the earth field is 0.042 Tesla.
​

Answer 1

Answer:

I really don't know this ans

sorry

Answer 2

Answer:4.3×10^-5

Explanation:the rule used to solve this is =(-n×Δ∅)÷Δt     ,  n stands for number of turns                      ∅ stands for magnetic flux , ∅max=area×β        ∅is max whenβ is horizontal(β in this question =0.042)        t stands for time

at first the thing which moved from east to west is the seconds arm          making a semi circle shape , the lenth of the sec arm represents radius of this semi circle                 now we have the area = 0.5×(14×10-²  )×π                     and also the number of turns which is =0.5        and also β                the seconds arm indicated time in seconds , moving half a rotation(at first it was pointing at 3 oclock then it pointed at 9oclock) that equals=Δt= 30 seconds                   ,                                      when the sec arm was at east(3 oclock) represents ∅1 which equals= ∅max=area ×β                 the sec arm at west(9oclock) this position represents ∅2         when the sec arm went at west it  became in the opposite direction of its older location so -∅2=∅1          ∅2 =-∅max=-1×area×β                Δ∅ =∅2-∅1                               thats it                                     ?سوال ل صاحب السوال اخبارك بعد سنة