Question

A helium nucleus is completing one round of a circle of radius 0.8 m in 2 seconds. Find the magnetic field induction at the centre of the circle

Answer 1

$<b><i>$The expression for the magnetic field for an electron revolving around a nucleus is given as

$B = \frac{μ_o}{4\pi} \frac{2\: \pi \: NI}{r}$

As the number of rotation is 1 there we have,

$B \: = \frac{μ_o}{4\pi} \frac{2πI}{r}$

$B \: = \frac{μ_o}{2r} I$
we know that current is defined as charge per unit time (I=q/t) and for helium atom there are only two charges therefore the current for helium nucleus is given as following:-

=>I = 2e/t

On substituting the above equation in magnetic field equation we get

$B \: = \frac{μ_o}{2r} \frac{2e}{t}$

$B \: = \frac{μ_o}{r} \frac{e}{t}$

On substituting the values we get

$B = \frac{1.67 \times {10}^{ - 19}C }{0.8m \times 2s} \: μ_o \: B = 1 \times {10}^{ - 19} \times μ_o$

Thus, the magnetic field at the centre of the circle is 

$B= {10^{} }^{ - 19} ×μ_o$

Thanks

Have a colossal day ahead..

Be Brainly