NEET Class neet Answered
please answer this
Asked by Prashant DIGHE | 06 Dec, 2019, 10:56: PM
Expert Answer
Left side figure shows the current loop made up of circular arc and chord of the same circle. Angle subtended by chord and arc is θ.
To get magnetic induction at P due to current i passing through loop as shown in figure,
let us calculate the magnetic induction separately for arc and chord.
Figure given in the middle helps us to get magentic induction B1 at P.
Let us consider small current element idl which is at a distance l from the foot of perpendicular drawn from P to chord.
Using Biot-Severt's law, we get, magnetic induction dB due to this current element as
dB = ( μo / 4π ) idl sinφ / ( l2 + d2 ) = ( μo / 4π ) idl d / ( l2 + d2 )3/2 ...................(1)
let us use the substitution l = d tanα ( see figure )
Hence dl = d sec2α dα
Using these substitutions, eqn.(1) becomes, dB = ( μo i / 4π d) cosα dα ................(2)
We get magnetic induction B1 by integrating eqn.(2) with limits from -θ/2 to +θ/2
B1 = ( μo i / 2π d) sin(θ/2) .....................(3)
since d = R cos(θ/2) , eqn,(3) becomes, B1 = ( μo i / 2π R) tan(θ/2) ...................(4)1
Direction of this magnetic induction is inward ( please check using right hand rule )
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Right side figure helps us to get magnetic induction at P due to circular arc.
Let us consider current element idl in the circular arc. Magnetic induction dB at P is given by
dB = ( μo / 4π ) ( idl / R2 )
Magnetic induction B2 due to full arc will be sum of above dB over arc length Rθ
Hence magnetic induction B2 = ( μo i / 2πR ) ( θ/2 )
Direction of this magnetic induction is outward
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net magnetic induction due to loop = B1 - B2 = ( μo i / 2π R) [ tan(θ/2) - ( θ/2 ) ]
when θ < π , [ tan(θ/2) - ( θ/2 ) ] > 0 , hence direction of net magnetic induction due to loop is inward
when θ > π , [ tan(θ/2) - ( θ/2 ) ] < 0 , hence direction of net magnetic induction due to loop is outwar
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Answer to the question :- option (3) inward as long as θ < π
Answered by Thiyagarajan K | 08 Dec, 2019, 08:49: PM
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