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a long wire carrying a current i is bent form find d magnetic field B at a point on d bisector of dis angl situated at a distance x from d vertexa long st wire of radius r carries a current i & is placed horizontally in a uniform magnetic field B pointing vertically upward d current is uniformly distributed over its cross section at wat points vil d resultant magnetic field have max magnitude ? wat vil b d maxmagnitude ? wat vil b d min magnitude of d resultant magnetic field?2 maro iz ekzam pilz yelp

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
ROSHAN MUJEEB

Last Activity: 4 Years ago

θ1​,θ2​with the end of wire is written as
B=4πaμ0​I​[sinθ1​+sinθ2​]

In this case :
a=sin(2θ​)θ1​=2θ​andθ2​=90​
therefore we get magnetic field due to one wire
Blower​=4πaμ0​I​⎣⎢⎢⎢⎡​xsin(2θ​)1+sin(2θ​)​⎦⎥⎥⎥⎤​
Bupper​also same
therefore Net field
Bnet​=Blower​+Bupper​
implies that2×4πμ0​I​⎣⎢⎢⎢⎡​xsin(2θ​)1+sin(2θ​)​⎦⎥⎥⎥⎤​
2πxμ0​I​⎣⎢⎢⎢⎡​sin(2θ​)1+sin(2θ​)​⎦⎥⎥⎥⎤​2πxμ0​I​⎣⎢⎢⎢⎢⎡​2sin(4θ​)cos(4θ​){sin(4θ​)+cos(4θ​)}2​⎦⎥⎥⎥⎥⎤​​
Now dividing numerator and dinominator bycos2(4θ​)
4πxμ0​I​⎣⎢⎢⎢⎢⎡​tan(4θ​){1+tan(4θ​)}2​⎦⎥⎥⎥⎥⎤​4πxμ0​I​[{1+tan(4θ​)}2]cot(4θ​)​
Given
B=Kcot(4θ​)
And Putting the value from above expression
K=4πxμ0​I​[{1+tan(4θ​)}2]

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