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A long straight wire, carrying current I...

A long straight wire, carrying current I, is bent at its midpoint to form an angle of `45^@`. Find the induction of magnetic field at point P, distant R from the point of bending (as shown in)

Text Solution

Verified by Experts

Since point P lies on axis of
straight part ab, therefore , magnetic induction due to this part is
equal to zero.
For part bc: From fig
`r=Rcos 45^@`.
Since both the ends b and c are
on the same side of normal PN,
therefore, `alpha` is negative and `beta` is positive.
Hence, `alpha=-45^@ and beta=+90^@`.
Using `B=(mu_0I)/(4pir)(sin alpha+sin beta)`, we have `B=((sqrt2-1)mu_0I)/(4piR)`
(into the plane of paper)
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Knowledge Check

  • A long stragith wire, carrying a current l is bent at its mid point to form an angle 60^(0) . AT a point P, distance R from the point of bending the magnetic field is

    A
    `((sqrt(2)-1)mu_(0)i)/(4piR)`
    B
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    C
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    D
    `(mu_(0)i)/(8R)`

  • A long wire carrying a current i is bent to form a plane angle. The magnitude of magnetic field at height d , above the point of bending

    A
    `(sqrt(2) mu_(0)i)/(4pid)`
    B
    `(3sqrt(2) mu_(0)i)/(4pid)`
    C
    `(5sqrt(2) mu_(0)i)/(4pid)`
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  • A straight wire of finite length carrying current I subtends an angle of 60^(@) at point P as shown. The magnetic field at P is

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    C
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    D
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