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A proton of mass m and charge q enters a magnetic field B with a velocity v at an angle θ with the  direction of B. The radius of curvature of the resulting path is

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Important Questions on Magnetic Effect of Current

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Two particles A and B of masses mA and mB, respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB, respectively and the trajectories are as shown in the figure. Then

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A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a 
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An electron moves with a velocity 1×103 m s-1 in a magnetic field of induction 0.3 T at an angle 30°. If em of electron is 1.76×1011C kg-1, the radius of the path is nearly
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A charged particle of charge q and mass m enters perpendicularly in a magnetic field B. Kinetic energy of the particle is E; then frequency of rotation is
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A beam of particles with specific charge 108 C kg-1 is entering with velocity 3×105 m s-1 by making an angle of 30° with the uniform magnetic field of 0.3 T. Radius of curvature of path of particle is
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If an electron enters a magnetic field with its velocity pointing in the same direction as the magnetic field, then
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A charge of 1 C is moving in a perpendicular magnetic field of 0.5 T with a velocity of 10 m s-1. Force experienced by the charge is 
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An electron accelerated by 200 V, enters a magnetic field. If its velocity is 8.4×106 m s-1. then em for it will be (in C kg-1)