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Q.

The magnetic field $B$ due to a current carrying circular loop of radius $12 c m$ at its center is $0.5 \times 10^{- 4} T$ . Find the magnetic field due to this loop at a point on the axis at a distance of $5.0 c m$ from the center.

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a

$B_{2} = - 3.9 \times 10^{- 5} T$

b

$B_{2} = 3.9 \times 10^{- 5} T$

c

$B_{2} = 3.09 \times 10^{- 5} T$

d

$B_{2} = - 3.009 \times 10^{- 5} T$

answer is B.

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Detailed Solution

$B_{1} = \frac{μ_{0} i}{2 R}$

and that at an axial point, $B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + x^{2})}^{\frac{3}{2}}}$

Thus, we can write

$\frac{B_{2}}{B_{1}} = \frac{R^{3}}{{(R^{2} + x^{2})}^{\frac{3}{2}}} B_{2} = B_{1} [\frac{R^{3}}{{(R^{2} + x^{2})}^{\frac{3}{2}}}]$

Substituting the values,

$B_{2} = (0.5 \times 10^{- 4}) (\frac{{(12)}^{3}}{{(144 + 25)}^{\frac{3}{2}}}) = 3.9 \times 10^{- 5} T$

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