A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched OFF and the same magnetic field is maintained, the electrons move
A.
in an elliptical orbit
B.
in a circular orbit
C.
along a parabolic path
D.
along a straight line
Answer :
in a circular orbit
Solution :
If both electric and magnetic fields are present and are perpendicular to each other and the particle is moving perpendicular to both of them with $${F_e} = {F_m}.$$ In this situation $$E \ne 0$$ and $$B \ne 0$$ and $${F_e} + {F_m} = 0.$$
But if electric field becomes zero, then only force due to magnetic field exists. And $$E$$ is perpendicular to the $$B$$ so the charge moves along a circle.
Releted MCQ Question on Electrostatics and Magnetism >> Magnetic Effect of Current
Releted Question 1
A conducting circular loop of radius $$r$$ carries a constant current $$i.$$ It is placed in a uniform magnetic field $${{\vec B}_0}$$ such that $${{\vec B}_0}$$ is perpendicular to the plane of the loop. The magnetic force acting on the loop is
A battery is connected between two points $$A$$ and $$B$$ on the circumference of a uniform conducting ring of radius $$r$$ and resistance $$R.$$ One of the arcs $$AB$$ of the ring subtends an angle $$\theta $$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is
A.
proportional to $$2\left( {{{180}^ \circ } - \theta } \right)$$
A proton, a deuteron and an $$\alpha - $$ particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $${r_p},{r_d},$$ and $${r_\alpha }$$ denote respectively the radii of the trajectories of these particles, then
A circular loop of radius $$R,$$ carrying current $$I,$$ lies in $$x - y$$ plane with its centre at origin. The total magnetic flux through $$x - y$$ plane is