Question: Q. 1. (i) Write an expression for the force $\vec{F}$ acting on a particle of mass $m$ and charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$. Under what conditions will it move in (a) a circular path and (b) a helical path?
(ii) Show that kinetic energy of the particle moving in magnetic field remains constant.
U] [Delhi III 2017]
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Solution:
Ans. (i) Expression for force acting on charged particle 1
(a) Condition for circular path
(b) Condition for helical path
(ii) Showing Kinetic energy is constant
(i) $\vec{F}=q(\vec{v} \times \vec{B})$
(a) When velocity of charge particle and magnetic field are perpendicular to each othe
(b) When velocity is neither parallel norperpendicular to the magnetic field.
$1 / 2$
(ii) The force, experienced by the charged particle, is perpendicular to the instantaneous velocity $\vec{v}$, at all instants. Hence the magnetic force cannot bring any change in the speed of the charged particle. Since speed remains constant, the kinetic energy also stays constant.
[CBSE Marking Scheme 2017]
[II Q. 2. State the Lorentz’s force and express it in vector form. Which pair of vectors are always perpendicular to each other? Derive the expression for the force acting on a current carrying conductor of length $L$ in a uniform magnetic field ’ $B$ ‘.
U [Delhi Comptt. I 2017]
Ans. Lorentz force
Expression in vector form $1 / 2$ Identification of pair of vectors Derivation of expression of force $1 / 2$ Lorentz magnetic force is force experienced by a charged particle of charge ’ $q$ ’ moving in magnetic field $\vec{B}$ with $\vec{v}$
$$ \overrightarrow{F_{m}}=q(\vec{v} \times \vec{B}) $$
$1 / 2$
[CBSE Marking Scheme, 2017]
