Question

# A helium nucleus (charge $$+2e$$) completes one round of a circle of radius $$0.8 m$$ in $$2sec$$. Find the magnetic field at the centre of the circle.

Solution

## The expression of magnetic field for an electron in a circular path isB $$=\dfrac { { \mu }_{ o }2\pi NI }{ 4\pi r } \\$$where, I= current, N= no. of rotation (Here in this case)= 1B $$=\dfrac { { \mu }_{ o }2\pi I }{ 4\pi r } \\ =\dfrac { { \mu }_{ o }I }{ 2r } \\ \\$$We know that, Helium has 2 charges. And the formula of current I$$=\dfrac { 2e }{ t }$$Also magnetic field B $$=\dfrac { { \mu }_{ 0 }2e }{ 2rt } =\dfrac { { \mu }_{ 0 }2e }{ rt } \\ =(\frac { 1.67\times { 10 }^{ -19 } }{ 0.8\times 2 } )\times 4\pi \times { 10 }^{ -7 }\\ =12.56\times { 10 }^{ -26 }T$$Physics

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