CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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 is
B $$=\dfrac { { \mu  }_{ o }2\pi NI }{ 4\pi r } \\ $$
where, I= current, N= no. of rotation (Here in this case)= 1
B $$=\dfrac { { \mu  }_{ o }2\pi I }{ 4\pi r } \\ =\dfrac { { \mu  }_{ o }I }{ 2r } \\ \\ $$

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image