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CBSE Class 12 Physics 2019 Delhi Set 3 Paper

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Question : 4 of 8
Marks: +1, -0
A charged particle qq is moving in the presence of a magnetic field BB which is inclined to an angle 3030^{\circ} with the direction of motion of the particle. Draw the trajectory followed by the particle in the presence of the field and explain how the particle describes this path.
Solution:  
Since magnetic force is given as,
F=q(v×B)\overset{\rightarrow}{F}=q\left(\overset{\rightarrow}{v}\times \overset{\rightarrow}{B}\right)
 or F=qvBsinθ\text{ or } F = q v B \sin \theta
 here, θ=30\text{ here, } \theta = 30^{\circ}
F=qvB2F = \frac{q v B}{2}
Velocity component parallel to the field moves particle in straight line and velocity component perpendicular to the field creates circular motion thus the combined effect is the helical path.
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