A particle of charge $$-16$$ $$x10-18C$$ moving with velocity $$10$$ $$m/s$$ along the $$X-axis$$ enters a region where a magnetic field of induction B is along $$Y-axis$$ and electric field of magnitude $$104$$ $$V/m$$ is a. long the negative $$Z-axis$$. If the charged particle continues moving along the $$X-axis$$, the magnitude of B is

Question

Infinity Learn · Accepted Answer

The situation is shown in fig .We know that,$$F=$$ q (E+$$ V x $$B)or$$ F $$=$$ qE $$+$$ q $$(v$$ x $$B)Now$$ Fe $$=$$ $$gE=$$−$$16$$×$$10$$−$$18$$×$$104($$−$$k)=16$$×$$10$$−$$14kAnd$$ $$Fm=$$−$$16$$×$$10$$−$$18(10i$$×$$Bj)=$$−$$16$$×$$1017B(k)∴ $$F=Fe+Fm=16$$×$$10$$−$$14k$$−$$16$$×$$10$$−$$17BkSince$$, the particle will continue to move along $$+X-axis$$, so resultant force is zero. Therefore, Fe $$+$$ Fm $$=$$ $$0or$$ $$16$$×$$10$$−$$14k=16$$×$$10$$−$$17Bk$$∴ $$B=103Wb/m2$$

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