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The magnetic moment of a current carrying loop is `2.1xx10^(-25) ampxxm^(2)`. The magnetic field at a point on its axis at a distance of `1 Å` isA. `4.2xx10^(-2)weber//m^(2)`B. `4.2xx10^(-3)weber//m^(2)`C. `4.2xx10^(-4)weber//m^(2)`D. `4.2xx10^(-5)weber//m^(2)` |
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Answer» Correct Answer - A Field at a point `x` from the centre of a current carrying loop on the axis is `B=(mu_(0))/(4pi).(2M)/(x^(3))=(10^(-7)xx2xx2.1xx10^(-25))/(10^(-10))^(3)` `4.2xx10^(-32)xx10^(30)=4.2xx10^(-2) W//m^(2)` |
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