The magnetic moment of a current carrying loop is 2.1 × 10−25 Am2. The magnetic field at a point on its axis at a distance of 1Ao is

Field at a point x from the centre of a current carrying loop on the axis isB = μ04π.2Mx3= 10−7 × 2 × 2.1 × 10−2510−103 =4.2 × 10−32 × 1030= 4.2 × 10−2 Wb/m2

The magnetic moment of a current carrying loop is 2.1 × 10−25 Am2. The magnetic field at a point on its axis at a distance of 1Ao is

Field at a point x from the centre of a current carrying loop on the axis isB = μ04π.2Mx3= 10−7 × 2 × 2.1 × 10−2510−103 =4.2 × 10−32 × 1030= 4.2 × 10−2 Wb/m2

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The magnetic moment of a current carrying loop is $2 .1 \times 10^{- 25} {Am}^{2}$ . The magnetic field at a point on its axis at a distance of $1 \overset{o}{A}$ is

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$4 .2 \times 10^{- 2} Wb / m^{2}$

$4 .2 \times 10^{- 3} Wb / m^{2}$

$4 .2 \times 10^{- 4} Wb / m^{2}$

$4 .2 \times 10^{- 5} Wb / m^{2}$

answer is A.

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Detailed Solution

Field at a point x from the centre of a current carrying loop on the axis is

$\begin{array}{l} B = \frac{μ_{0}}{4 π} . \frac{2 M}{x^{3}} = \frac{10^{- 7} \times 2 \times 2 .1 \times 10^{- 25}}{{(10^{- 10})}^{3}} \\ = 4 .2 \times 10^{- 32} \times 10^{30} = 4 .2 \times 10^{- 2} Wb / m^{2} \end{array}$