(D) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path with a radius $r$ given by $r = \

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(D) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path with a radius $r$ given by $r = \frac{mv}{qB}$.For the particle not to strike the opposite plate,the radius of its circular path must be less than the distance $d$ between the plates.Therefore,the condition is $r Substituting the expression for $r$,we get $\frac{mv}{qB} Rearranging this inequality to solve for $B$,we obtain $B > \frac{mv}{qd}$.

Class 12 Physics Moving Charges and Magnetism Motion of Charged Particle In Magnetic Field

Medium

$A$ charged particle of mass $m$ and charge $q$ is projected with velocity $v$ into a region of uniform magnetic field $B$ confined between two parallel plates separated by a distance $d$,as shown in the figure. What is the condition for the charged particle not to strike the opposite plate?

A
$B > \frac{qd}{mv}$
B
$B < \frac{mv}{qd}$
C
$B < \frac{qd}{mv}$
D
$B > \frac{mv}{qd}$

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

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An electron projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$

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An electron $(e^-)$ is moving in the North direction and the magnetic field at this place is in the upward direction. Then,the electron $(e^-)$ will be deviated towards:

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An electron having charge $1.6 \times 10^{-19} \, C$ and mass $9 \times 10^{-31} \, kg$ is moving with $4 \times 10^6 \, m/s$ speed in a magnetic field of $2 \times 10^{-1} \, T$ in a circular orbit. The force acting on the electron and the radius of the circular orbit are:

Medium

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$A$ charged particle of mass $m$ and charge $q$ is projected with velocity $v$ into a region of uniform magnetic field $B$ confined between two parallel plates separated by a distance $d$,as shown in the figure. What is the condition for the charged particle not to strike the opposite plate?

Similar Questions

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $times$)

An electron projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$

An electron $(e^-)$ is moving in the North direction and the magnetic field at this place is in the upward direction. Then,the electron $(e^-)$ will be deviated towards:

$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

An electron having charge $1.6 \times 10^{-19} \, C$ and mass $9 \times 10^{-31} \, kg$ is moving with $4 \times 10^6 \, m/s$ speed in a magnetic field of $2 \times 10^{-1} \, T$ in a circular orbit. The force acting on the electron and the radius of the circular orbit are:

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