Question

an electron revolves in a circle of radius 0.4 angstroms with a speed of 10 power 4 ms power minus1. the magnitude of the magnetic field at the centre of the circular path due to the motion of the electron is​

Answer 1

Question :

An electron revolves in a circle of radius 0.4 A° with a speed of 10⁴ ms⁻¹. The magnitude of the magnetic field at the centre of the circular path due to the motion of the electron is​

Answer :

=> B = 0.1 Wb/m²

Explanation :

we know,

   Magnetic field at the centre,

                $\boxed{\it B=\frac{\mu_0i}{2r} }$

where,

B - magnitude of magnetic field

i - current

r - radius of the circle

    $\boxed{i=\frac{q}{t}} \ \ \& \ \ \boxed{t=\frac{2 \pi r}{v} }$

         $\implies \bf B=\frac{\mu_0q}{2rt} \\\\\\ \implies \bf B=\frac{\mu_0q \times v}{2r \times 2\pi r} \\\\\\ \boxed{\bf \implies B=\frac{\mu_0qv}{4\pi r^2} }$

q - charge of the electron

v - speed of the electron

μ₀ = 4π × 10⁻⁷ H/m

q  = 1.6 × 10⁻¹⁹ C

r   = 0.4 A°

   = 0.4 × 10⁻¹⁰ m           [ 1 A° = 10⁻¹⁰ m ]

v  = 10⁴ m/s

Substitute the above values,

         $B=\frac{4\pi \times 10^{-7} \times 1.6 \times 10^{-19} \times 10^4}{4\pi \times (0.4\times 10^{-10})^2} \\\\ \\ B=\frac{1.6 \times 10^{-22}}{0.16 \times 10^{-20}} \\\\\\ B=\frac{16 \times 10^{-23}}{16 \times 10^{-22}} \\\\\\ B=10^{-1} \\\\\\\ B=0.1 \ Wb/m^2$

the magnetic field at the centre of the circular path due to the motion of the electron is​ 0.1 Wb/m²