First slide

Magnetic field due to a current carrying circular ring

Question

An electron is revolving around a proton in a circular orbit of diameter 1 $Å$ . If it produces a magnetic field of 14 weber/m² at the proton, then its angular velocity will be about

Easy

A

4x 10¹⁶ rad,/sec
B

10¹⁶ rad/sec
C

4x 10¹⁵ rad,sec
D

10¹⁵ rad/sec

Solution

$\begin{array}{l} B = \frac{μ_{0}}{4 π} \times \frac{2 πi}{r} = \frac{μ_{0}}{4 π} \times 2 π (\frac{e}{t}) \times \frac{1}{r} \\ = \frac{μ_{0}}{4 π} \times 2 π \times \frac{e}{(2 πr / v)} \times \frac{1}{r} \\ = \frac{μ_{0}}{4 π} \times 2 π \times \frac{ev}{2 πr} \times \frac{1}{r} = \frac{μ_{0}}{4 π} \times \frac{eω}{r} \\ ∴ ω = \frac{4 π}{μ_{0}} \times \frac{rB}{e} = 10^{7} \times \frac{10^{- 10} \times 14}{2 \times 1 \cdot 6 \times 10^{- 19}} \\ = 4 \times 10^{16} rad / \sec \end{array}$

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