A charged particle moving with a velocity vec | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic force is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since $\vec{F}$ is perpendicular to $\vec{v}$,their dot product must be zero:

Vedclass · Accepted Answer

$\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$ with $\frac{v_1}{v_2} = \frac{B_1}{B_2}$

Vedclass · Answer

(B) The magnetic force is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since $\vec{F}$ is perpendicular to $\vec{v}$,their dot product must be zero: $\vec{F} \cdot \vec{v} = 0$.$(F_1 \hat{i} + F_2 \hat{j}) \cdot (v_1 \hat{i} + v_2 \hat{j}) = F_1 v_1 + F_2 v_2 = 0 \Rightarrow \frac{F_1}{F_2} = -\frac{v_2}{v_1} \quad (I)$.Also,$\vec{F}$ is perpendicular to $\vec{B}$,so $\vec{F} \cdot \vec{B} = 0$.Let $\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$. Then $(F_1 \hat{i} + F_2 \hat{j}) \cdot (B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}) = F_1 B_1 + F_2 B_2 = 0 \Rightarrow \frac{F_1}{F_2} = -\frac{B_2}{B_1} \quad (II)$.Comparing $(I)$ and $(II)$,we get $\frac{v_2}{v_1} = \frac{B_2}{B_1}$,which implies $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.Thus,$\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$ satisfies the condition $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.

$A$ charged particle moving with a velocity $\vec{v} = v_1 \hat{i} + v_2 \hat{j}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F} = F_1 \hat{i} + F_2 \hat{j}$. Here $v_1, v_2, F_1, F_2$ are all constants. Then $\vec{B}$ can be

Similar Questions

$A$ positively charged $(+q)$ particle of mass $m$ has kinetic energy $K$ and enters vertically downward into a horizontal magnetic field of induction $\vec{B}$. The acceleration of the particle is:

The radius of curvature of the path of a charged particle moving in a static uniform magnetic field is

The distance between two plates is $1 \ cm$ and the potential difference is $1000 \ V$. $A$ magnetic field $B = 1 \ T$ is applied. If an electron passes through without any deflection,what will be its velocity?

$A$ charge $q$ is moving in a magnetic field. Then,the magnetic force does not depend upon:

Vedclass Test Series

Exam Paper Generator

Online Exam Module