(D) When a charged particle enters a perpendicular magnetic field,it follows a circular path. The time period of a full circular orbit is $T = \frac{2

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$T \left( \frac{\pi - 2\theta}{2\pi} \right)$

(D) When a charged particle enters a perpendicular magnetic field,it follows a circular path. The time period of a full circular orbit is $T = \frac{2\pi m}{QB}$.From the geometry of the path,the particle enters the field at an angle $\theta$ with the normal to the boundary $YZ$. The angle subtended by the arc of the circular path inside the magnetic field at the center is $\phi = \pi - 2\theta$.The time spent in the magnetic field is given by $t = \frac{\phi}{2\pi} T$.Substituting $\phi = \pi - 2\theta$,we get $t = \frac{\pi - 2\theta}{2\pi} T = T \left( \frac{\pi - 2\theta}{2\pi} \right)$.

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

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

A
$T\theta$
B
$2T\theta$
C
$T \left( \frac{\pi + 2\theta}{2\pi} \right)$
D
$T \left( \frac{\pi - 2\theta}{2\pi} \right)$

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

$A$ particle of mass $0.6\, g$ and having a charge of $25\, nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4\, m/s$ in a uniform magnetic field. If the particle continues to move with uniform velocity,the value of the magnetic induction is $(g = 10\, m/s^2)$.

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$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

Medium

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$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

EasyAP EAMCET 2002

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$A$ charge having $q/m$ equal to $10^8 \, C/kg$ and with velocity $3 \times 10^5 \, m/s$ enters into a uniform magnetic field $0.3 \, T$ at an angle $30^{\circ}$ with the direction of the field. The radius of curvature will be ...... $cm$.

MediumAIPMT 2000

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$A$ block of mass $m$ and charge $q$ is released on a long smooth inclined plane. $A$ magnetic field $B$ is constant,uniform,horizontal,and parallel to the surface as shown. Find the time from the start when the block loses contact with the surface.

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$A$ particle of mass $0.6\, g$ and having a charge of $25\, nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4\, m/s$ in a uniform magnetic field. If the particle continues to move with uniform velocity,the value of the magnetic induction is $(g = 10\, m/s^2)$.

$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

$A$ charge having $q/m$ equal to $10^8 \, C/kg$ and with velocity $3 \times 10^5 \, m/s$ enters into a uniform magnetic field $0.3 \, T$ at an angle $30^{\circ}$ with the direction of the field. The radius of curvature will be ...... $cm$.

$A$ block of mass $m$ and charge $q$ is released on a long smooth inclined plane. $A$ magnetic field $B$ is constant,uniform,horizontal,and parallel to the surface as shown. Find the time from the start when the block loses contact with the surface.

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