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The magnetic field at O due to current i...

The magnetic field at O due to current in the infinite wire forming a loop as a shown in Fig.

A

`(mu_0I)/(2pid) (cos phi_1+cos phi_2)`

B

`(mu_0)/(4pi) (2I)/d (tan theta_1+tan theta_2)`

C

`(mu_0)/(4pi) I/d (sin phi_1+sin phi_2)`

D

`(mu_0)/(4pi) I/d (cos theta_1+cos theta_2)`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) Using `B=(mu_0i)/(4pia)[sin theta_1+sin theta_2]`
But ` theta_1+phi_1=90^@ or theta_1=90^@-phi,`
`sin theta_1= sin(90^@-phi_1)=cos phi_1`
Similarly, `sin theta_2=cos phi_2`
`B_(n et)=(mu_0I)/(2pia)(cos phi_1+cos phi_2)`
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Knowledge Check

  • The magnetic field at the origin due to the current flowing in the wire is

    A
    `-(mu_0I)/(8pia)(hati+hatk)`
    B
    `(mu_0I)/(2pia)(hati+hatk)`
    C
    `(mu_0I)/(8pia)(-hati+hatk)`
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  • Net magnetic field at the centre of the circle O due to a currying loop as shown in figure is (thetalt180^(@)) .

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    B
    perpendicular to paper inwards
    C
    perpendicular to paper outwards
    D
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  • Magnetic field produced at the point O due to current flowing in an infinite wire shaped as shown in the figure is

    A
    `(mu_(0)I)/(4piR)`
    B
    `(mu_(0)I)/(4R)-(mu_(0)I)/(4piR)`
    C
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    D
    `(mu_(0)I)/(4R)+(mu_(0)I)/(4piR)`

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