(D) The charge $q = 1000e = 1000 \times 1.6 \times 10^{-19} \ C = 1.6 \times 10^{-16} \ C$. The frequency of rotation $f = 1 \ Hz$. The equivalent cur

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(D) The charge $q = 1000e = 1000 \times 1.6 \times 10^{-19} \ C = 1.6 \times 10^{-16} \ C$. The frequency of rotation $f = 1 \ Hz$. The equivalent current $I = qf = 1.6 \times 10^{-16} \times 1 = 1.6 \times 10^{-16} \ A$. The magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2r}$. Given $B = x \mu_0$,where $x = 2 \times 10^{-16}$. So,$x \mu_0 = \frac{\mu_0 I}{2r} \implies x = \frac{I}{2r}$. Substituting the values: $2 \times 10^{-16} = \frac{1.6 \times 10^{-16}}{2r}$. $2 = \frac{0.8}{r} \implies r = \frac{0.8}{2} = 0.4 \ m$.

Class 12 Physics Moving Charges and Magnetism Biot-Savart's Law and its application

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$A$ particle carrying a charge equal to $1000$ times the charge on an electron is rotating at $1$ rotation per second in a circular path of radius $r \ m$. If the magnetic field produced at the centre of the path is $x$ times the permeability of vacuum $\mu_0$,the radius $r$ in $m$ is: $[e = 1.6 \times 10^{-19} \ C], [x = 2 \times 10^{-16}]$

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$0.04$
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$0.02$
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$0.4$

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Moving Charges and Magnetism

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Biot-Savart's Law and its application

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