(A) The magnetic field is $\vec{B} = B\hat{j}$. The initial velocity is $\vec{v} = v \cos \alpha \hat{j} + v \sin \alpha \hat{k}$.The magnetic force o

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its $x-$coordinate can never be positive

(A) The magnetic field is $\vec{B} = B\hat{j}$. The initial velocity is $\vec{v} = v \cos \alpha \hat{j} + v \sin \alpha \hat{k}$.The magnetic force on the proton is $\vec{F} = q(\vec{v} \times \vec{B}) = q[(v \cos \alpha \hat{j} + v \sin \alpha \hat{k}) \times B\hat{j}] = qvB \sin \alpha (\hat{k} \times \hat{j}) = -qvB \sin \alpha \hat{i}$.Since the force is in the negative $x-$direction,the particle will accelerate in the negative $x-$direction. Thus,the $x-$coordinate will always be negative (or zero at the origin). Therefore,the $x-$coordinate can never be positive.

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

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In a region of space,a uniform magnetic field $B$ exists in the $y-$direction. $A$ proton is fired from the origin,with its initial velocity $v$ making a small angle $\alpha$ with the $y-$direction in the $yz$ plane. In the subsequent motion of the proton,

A
its $x-$coordinate can never be positive
B
its $x-$ and $z-$ coordinates cannot both be zero at the same time
C
its $z-$ coordinate can never be negative
D
its $y-$ coordinate will be proportional to the square of its time of flight

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The particle leaves the magnetic field at point $D$. Calculate the distance $CD$. (Assume the velocity of the particle $v = 10^7 \, m/s$)

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$A$ beam of protons enters a uniform magnetic field of $0.314 \ T$ with a velocity $4 \times 10^5 \ ms^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field. The path of the beam is (mass of proton $= 1.6 \times 10^{-27} \ kg$).

EasyAP EAMCET 2018

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$A$ charged particle with specific charge $S$ moves undeflected through a region of space containing mutually perpendicular uniform electric and magnetic fields $E$ and $B$. When the electric field is switched off,the particle will move in a circular path of radius:

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$A$ positive charge particle having charge $q$ and mass $m$ has velocity $\vec v = v\left( {\frac{{\hat i + \hat k}}{{\sqrt 2 }}} \right)$ in the magnetic field $\vec B = B\hat j$ and electric field $\vec E = E\hat j$ at the origin. Its speed as a function of $y$ is:

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What is the radius of the path of an electron (mass $m = 9 \times 10^{-31} \; kg$ and charge $q = 1.6 \times 10^{-19} \; C$) moving at a speed of $v = 3 \times 10^{7} \; m/s$ in a magnetic field of $B = 6 \times 10^{-4} \; T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. (Given: $1 \; eV = 1.6 \times 10^{-19} \; J$)

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In a region of space,a uniform magnetic field $B$ exists in the $y-$direction. $A$ proton is fired from the origin,with its initial velocity $v$ making a small angle $\alpha$ with the $y-$direction in the $yz$ plane. In the subsequent motion of the proton,

Similar Questions

$A$ beam of protons enters a uniform magnetic field of $0.314 \ T$ with a velocity $4 \times 10^5 \ ms^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field. The path of the beam is (mass of proton $= 1.6 \times 10^{-27} \ kg$).

$A$ charged particle with specific charge $S$ moves undeflected through a region of space containing mutually perpendicular uniform electric and magnetic fields $E$ and $B$. When the electric field is switched off,the particle will move in a circular path of radius:

$A$ positive charge particle having charge $q$ and mass $m$ has velocity $\vec v = v\left( {\frac{{\hat i + \hat k}}{{\sqrt 2 }}} \right)$ in the magnetic field $\vec B = B\hat j$ and electric field $\vec E = E\hat j$ at the origin. Its speed as a function of $y$ is:

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