1.

There is a circular loop of current of radius , as shown in figure: If current enters at point A and leaves at point B where angle AOB " is " 90^(@)and the wire of the loop is of uniform cross section area, then calculate the magnetic field intensity at the centre of the loop.

Answer»

0
`(3 mu_(0)i)/(16 r) `
`(mu_(0)i)/(32 r)`
`(3 mu_(0)i)/(32 r) `

Solution :LET R be the resistance of portion AB, then 3R will be the resistance of the remaining loop because AB is just a quarter of the circle. Both the parts will have the same potential difference between points A and B. For the same potential difference, using V=iR, we can understand that the current will be distributed inversely proportional to their resistances. Hence CURRENTS flowing in the two parts of the loop will be as follows:
`i_(1) = (3R)/(R + 3R) i = (3i)/4`
` i_(2) = R/(R+3R) i = i/4`
Magnetic field intensity at the centre due to part AB can be written as follows:
`B_(1) = 1/4 [(mu_(0)i_(1))/(2r)] = 1/4 [(mu_(0)(3i)/4)/(2r)] = (3 mu_(0)i)/(32r)`
Magnetic field intensity due to the part BCDA can be written as follows:
`B_(2) = 3/4 [(mu_(0)i_(1))/(2r)] = 3/4 [(mu_(0) i/4)/(2r)] = (3 mu_(0)i)/(32 r) `
Both the fields are having the same magnitude but the two are opposite in direction and hence the net magnetic field at the centre of the loop becomes zero.
Note that we can predict the answer with just a bit of mental calculations. One portion is a quarter of the circle but current is three times than the other part which is three QUARTERS of the circle. Hence, OPTION (a) is correct.


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