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What is the magnetic field strength at point a? B= ???? What is the direction of the magnetic field strength at point a? - to the left
What is the magnetic field strength at point a?
B= ????
What is the direction of the magnetic field strength at point a?
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- to the left. – out of page. – to the right. –into page - What is the magnetic field strength at point b? B = ???? -- What is the direction of the magnetic field strength at point b? - to the left. – out of page. – to the right. –into page -------------------------------------------------------------------------------- - What is the magnetic field strength at point c? B = ???? What is the direction of the magnetic field strength at point c? - to the left. – out of page. – to the right. –into page |
2.0 cm 4.0 cm 2.0 cm
Expert Solution
Magnetic field at a point due to a current carrying wire is given by :
B = uI/(2*pi*r)
where u = 1.26*10^-6
I = current
r = distance from the wire
So, magnetic field at a due to (upper wire), B1 = u*(I)/(2*pi*0.02) directed out of the plane
NOTE:--> distance of point a from upper wire = 2 cm = 0.02 m and I = 15 A
Similarly for lower wire, B2 = u*(I)/(2*pi*0.06) directed into the plane
So, net magnetic field at point a , Bnet = B1-B2 <----- as directions are opposite so subtraction
So, Bnet = u*(I)/(2*pi*0.02) - u*(I)/(2*pi*0.06)
= 1.26*10^-6*15/(2*pi*0.02) -1.26*10^-6*15/(2*pi*0.06)
= 1*10^-4 T <------------answer
Direction is out of the page as B1 is greater than B2 <---------answer
--------------------------------------------------------
For point b
B1 = 1.26*10^-6*15/(2*pi*0.02) into the page
B2 = 1.26*10^-6*15/(2*pi*0.02) into the page
So, Bnet = B1 + B2 <---- both are directed in same direction
So, Bnet = 3*10^-4 T <-----------answer
Direction = into the page <-----------answer
---------------------------------------------------------------
For point c,
Bnet = 1*10^-4 T
Direction = out of the page
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