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The magnetic field dvecB due to a small ...

The magnetic field `dvecB` due to a small current element `dvecl` at a distance `vecr` and element carrying current `i` is,

A

`dvecB=mu_(0)/(4pi)i((dveclxxvecr)/(r))`

B

`dvecB=mu_(0)/(4pi)i^(2)((dveclxxvecr)/(r))`

C

`dvecB=mu_(0)/(4pi)i^(2)((dveclxxvecr)/(r^(2)))`

D

`dvecB=mu_(0)/(4pi)i((dveclxxvecr)/(r^(3)))`

Text Solution

Verified by Experts

The correct Answer is:
D
`dB=(mu_(0))/(4pi).(i dl sintheta)/(r^(2))impliesdvecB=(mu_(0 ))/(4pi).(i(dveclxxvecr))/(r^(3))`
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Knowledge Check

  • The magnetic field bar(dB) due to a small current element bar(dl) at a distance vec(r) and carrying current 'I' is

    A
    `bar(dB)=(mu_(0))/(4pi)i((bar(dl)xxbar(r))/(r))`
    B
    `bar(dB)=(mu_(0))/(4pi)i^(2)((bar(dl)xxbar(r))/(r^(2)))`
    C
    `bar(dB)=(mu_(0))/(4pi)i^(2)((bar(dl)xxbar(r))/(r))`
    D
    `bar(dB)=(mu_(0))/(4pi)i((bar(dl)xxbar(r))/(r^(3)))`

  • The magnetic field vec(dB) due to a small current element vec(dl) carrying a current I at a distance vecr is given as

    A
    `vec(dB) = (mu_0)/(4pi) I[(vec(dI) xx vec(r))/(r)]`
    B
    `vec(dB) = (mu_0)/(4pi) I^2[(vec(dl) xx vec(r))/(r)]`
    C
    `vec(dB) = (mu_0)/(4pi) I^2[(vec(dl) xx vec(r))/(r^2)]`
    D
    `vec(dB) = (mu_0)/(4pi) I[(vec(dl) xx vec(r))/(r^3)]`

  • The magnetic field (d vecB) at a point due to an elemental conductor (d vecl) carrying a current (i) at a distance (vecr) from the elemetn, is given by

    A
    Faraday's Law
    B
    Columb's Law
    C
    Biot-Savart's Law
    D
    Ampere's Law
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