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Question: The magnetic moment of a current carrying loop is \(2.1 \times 10 ^ { - 25 } \mathrm { amp } \times ...

The magnetic moment of a current carrying loop is 2.1×10−25amp×m22.1 \times 10 ^ { - 25 } \mathrm { amp } \times \mathrm { m } ^ { 2 }. The magnetic field at a point on its axis at a distance of 1A1 A is

A

4.2×10−24.2 \times 10 ^ { - 2 } weber /m2/ \mathrm { m } ^ { 2 }

B

4.2×10−34.2 \times 10 ^ { - 3 } weber /m2/ \mathrm { m } ^ { 2 }

C

D

4.2×10−54.2 \times 10 ^ { - 5 } weber /m2/ \mathrm { m } ^ { 2 }

Answer

4.2×10−24.2 \times 10 ^ { - 2 } weber /m2/ \mathrm { m } ^ { 2 }

Explanation

Solution

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

B=μ04π⋅2Mx3=10−7×2×2.1×10−25(10−10)3\mathrm { B } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 M } { x ^ { 3 } } = \frac { 10 ^ { - 7 } \times 2 \times 2.1 \times 10 ^ { - 25 } } { \left( 10 ^ { - 10 } \right) ^ { 3 } }

=4.2×10−32×1030=4.2×10−2 W/m2= 4.2 \times 10 ^ { - 32 } \times 10 ^ { 30 } = 4.2 \times 10 ^ { - 2 } \mathrm {~W} / \mathrm { m } ^ { 2 }