Question

A Helium nucleus (charge +2e) completes one round of a circle 0.8 m in 2s. Find the magnetic field at the centre of the circle.

Answer 1

Given:

Charge on Helium nucleus (q) = +2e = $\rm 2 \times 1.6 \times 10^{-19}$ C

Radius of circular loop (r) = 0.8 m

Time taken for one complete revolution (t) = 2s

To Find:

Magnetic field at the centre of the circle (B)

Answer:

Magnetic field at the centre of circular loop is given as:

$\boxed{ \bf{B = \dfrac{\mu_o I}{2r}}}$

I → Current

$\rm \mu_o$ → Permeability of free space ($\rm 4\pi \times 10^{-7}$)

As we know rate of flow of electric charge is known as current i.e.

$\rm I = \dfrac{q}{t}$

So,

$\bf B = \dfrac{\mu_o q}{2rt}$

By substituting values in the equation we get:

$\rm \implies B = \dfrac{4\pi \times {10}^{ - 7} \times \cancel{ 2 }\times \cancel{1.6} \times {10}^{ - 19} }{ \cancel{2 }\times \cancel{ 0.8} \times \cancel{2}} \\ \\ \rm \implies B =4\pi \times {10}^{ - 7 - 19} \\ \\ \rm \implies B =4\pi \times {10}^{ - 26} \ T$

$\therefore$ Magnetic field at the centre of the circle (B) = $\rm 4\pi \times {10}^{ - 26} \ T$

Answer 2

B = μ0q/2rt

= 2μ0e/2rt

= μ0e/rt

= 4π × 10^-7 × 1.6 × 10^-19/0.8 × 2

= 12.56 × 10^-26 T