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Magnetic field due to a ring having n tu...

Magnetic field due to a ring having n turns at a distance `x` on its axis is proportional to (if `r =` radius of ring)

A

`r/((x^(2)+r^(2)))`

B

`(r^(2))/((x^(2)_r^(2))^(3//2))`

C

`(nr^(2))/((x^(2)+r^(2))^(3//2))`

D

`(n^(2)r^(2))/((x^(2)+r^(2))^(3//2))`

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • The field due to a magnet at a distance R~ from the centre of the magnet is proportional

    A
    `R^(2)`
    B
    `R^(3)`
    C
    `1//R^(2)`
    D
    `1//R^(3)`

  • The electric field strength due to a ring of radius R at a distance x from its centre on the axis of ring carrying charge Q is given by E = (1)/(4 pi epsilon_(0)) (Qx)/((R^(2) + x^(2))^(3//2)) At what distance from the centre will the electric field be maximum ?

    A
    `x = R`
    B
    `x = (R )/(2)`
    C
    `x = +- (R )/(sqrt(2))`
    D
    `x = sqrt(2) R`

  • A uniformly charged ring of radius R is rotated about its axis with constant linear speed v of each of its particle. The ratio of electric field to magnetic field at a point P on the axis of the ring distant x=R from centre of ring is ( c is speed of light )

    A
    `(c^(2))/(v)`
    B
    `(v^(2))/( c)`
    C
    `(v)/(c )`
    D
    `(c )/ (v)`

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