The magnetic field B due to a current-carrying circular loop of radius 12 cm at its centre is 0.50×10−4T. Find the magnetic field due to this loop at a point on the axis at a distance of 5.0 cm from the centre

The magnetic field at the centre of a circular loop is B0=μ0i2aAnd that at an axial point is B=μ0ia22a2+x23/2Thus, BB0=a2a2+x23/2Or, B=0.50×10−4T×(12cm)3144cm2+25cm23/2=3.9×10−5T

The magnetic field B due to a current-carrying circular loop of radius 12 cm at its centre is 0.50×10−4T. Find the magnetic field due to this loop at a point on the axis at a distance of 5.0 cm from the centre

The magnetic field at the centre of a circular loop is B0=μ0i2aAnd that at an axial point is B=μ0ia22a2+x23/2Thus, BB0=a2a2+x23/2Or, B=0.50×10−4T×(12cm)3144cm2+25cm23/2=3.9×10−5T

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The magnetic field B due to a current-carrying circular loop of radius 12 cm at its centre is $0 .50 \times 10^{- 4} T .$ Find the magnetic field due to this loop at a point on the axis at a distance of 5.0 cm from the centre

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$6 .0 \times 10^{- 5} T$

$5 .9 \times 10^{- 5} T$

$3 .9 \times 10^{- 5} T$

$4 .9 \times 10^{- 5} T$

answer is B.

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Detailed Solution

The magnetic field at the centre of a circular loop is $B_{0} = \frac{μ_{0} i}{2 a}$
And that at an axial point is $B = \frac{μ_{0} {ia}^{2}}{2 {(a^{2} + x^{2})}^{3 / 2}}$
Thus, $\frac{B}{B_{0}} = \frac{a^{2}}{{(a^{2} + x^{2})}^{3 / 2}}$
Or, $B = (0 .50 \times 10^{- 4} T) \times \frac{(12 cm)^{3}}{{(144 {cm}^{2} + 25 {cm}^{2})}^{3 / 2}} = 3 .9 \times 10^{- 5} T$