Question

In Bohr's model of hydrogen atom, the electron circulates round the nucleus in a path of radius $5\times {10}^{-11}$m at a frequency of $6.8\times {10}^{15}$ revolutions per second. The value of magnetic induction at the centre of the orbit is :

• A. $12.27T$
• B. $10.8T$
• C. $13.2T$
• D. $13.6T$

Answer 1

Answer: (D)

$13.6T$

$\left|B\right|$ =$\frac{{\mu }_{0}i}{2r}$

I=$\frac{dq}{dt}$

= $qf$

= $1.6\times {10}^{-19}\times 6.8\times {10}^{15}$

=$10.88\times {10}^{-4}$

$\therefore \left|B\right|=\frac{4\pi \times {10}^{-7}\times 10.88\times {10}^{-4}}{2\times 5\times {10}^{-11}}$

=$13.6T$