A permanent magnet has shape of a thin disc with magnetic moment along its axis the radius of the disc is 1 cm. The magnetic field due to it can be assumed to be that due to a molecular current I flowing along the rim of the disc. It $¯ B$ at a distance of 10 cm from its centre on the axis is 30 $μ$ T, find I (approximately).

1 k A

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0.5 k A

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2.5 k A

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3 k A

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Solution

The correct option is B 0.5 k A
$8 = \frac{10^{- 7} \times 2 \times 3.14 \times i \times (1 \times 10^{- 2})^{2}}{(10 \times 10^{- 2} + 1 \times 10^{- 2}) 3 / 2}$
$\Rightarrow 30.0 \times 10^{- 6} = \frac{6.28 \times 10^{- 1} \times i \times 10^{- 4}}{(11 \times 10)^{3}}$
$\Rightarrow i = 0.5 k A$

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The correct option is B 0.5 k A8=10−7×2×3.14×i×(1×10−2)2(10×10−2+1×10−2)3/2⇒30.0×10−6=6.28×10−1×i×10−4(11×10)3⇒i=0.5 kA

The correct option is B 0.5 k A
$8 = \frac{10^{- 7} \times 2 \times 3.14 \times i \times (1 \times 10^{- 2})^{2}}{(10 \times 10^{- 2} + 1 \times 10^{- 2}) 3 / 2}$
$\Rightarrow 30.0 \times 10^{- 6} = \frac{6.28 \times 10^{- 1} \times i \times 10^{- 4}}{(11 \times 10)^{3}}$
$\Rightarrow i = 0.5 k A$