A particle carrying a charge equal to $$100$$ times the charge on an electron is making one revolution per second in a circular path of radius $$0.8$$ m. The value of the magnetic field produced at the centre will be $$($$μ$$0$$ $$=$$ permeability for $$vacuum)$$

Question

A particle carrying a charge equal to $$100$$ times the charge on an electron is making one revolution per second in a circular path of radius $$0.8$$ m. The value of the magnetic field produced at the centre will be $$($$μ$$0$$ $$=$$ permeability for $$vacuum)$$

Infinity Learn · Accepted Answer

$$i=qt$$  $$=$$  $$100$$ ×  eBcentre  $$=$$  μ$$04$$π . $$2$$$$π$$ir  $$=$$  μ$$04$$π $$=$$ $$2$$π × $$100er=$$ μ$$0$$ ×  $$200$$ × $$1.6$$ × $$10$$−$$194$$ × $$0.8$$ $$=$$ $$10$$−$$17$$  μ$$0$$

A particle carrying a charge equal to 100 times the charge on an electron is making one revolution per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be ( $μ_{0}$ = permeability for vacuum)

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A particle carrying a charge equal to 100 times the charge on an electron is making one revolution per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be (μ0 = permeability for vacuum)

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A particle carrying a charge equal to 100 times the charge on an electron is making one revolution per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be ( $μ_{0}$ = permeability for vacuum)