(B) The radius $r$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula: $r = \frac{mv}{qB}$.Rearranging for $B$, we g

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(B) The radius $r$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula: $r = \frac{mv}{qB}$.Rearranging for $B$, we get: $B = \frac{mv}{qr}$.Given values:$m = 1.66 \times 10^{-27} \ kg$$v = 8 \times 10^5 \ m/s$$q = 1.6 \times 10^{-19} \ C$$r = 8.3 \ cm = 8.3 \times 10^{-2} \ m$Substituting these values into the formula:$B = \frac{(1.66 \times 10^{-27}) \times (8 \times 10^5)}{(1.6 \times 10^{-19}) \times (8.3 \times 10^{-2})}$$B = \frac{13.28 \times 10^{-22}}{13.28 \times 10^{-21}}$$B = 0.1 \ T = 100 \ mT$.Therefore, the magnitude of the magnetic field is $100 \ mT$.

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

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$A$ proton moving with a velocity of $8 \times 10^5 \ m/s$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3 \ cm$, then the magnitude of the magnetic field is (Charge of proton $= 1.6 \times 10^{-19} \ C$ and mass of the proton $= 1.66 \times 10^{-27} \ kg$) (in $mT$)

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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 revolves around the nucleus in a circular path with angular momentum $\vec{L}$. $A$ uniform magnetic field $\vec{B}$ is applied perpendicular to the plane of its orbit. If the electron experiences a torque $\vec{\tau}$, then

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$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$

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Two particles $A$ and $B$ having equal charges $+6\,C$,after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii $2\,cm$ and $3\,cm$ respectively. The ratio of mass of $A$ to that of $B$ is

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An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

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$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. If the particle moves in a straight line, the value of the magnetic induction is $(g=10 \,ms^{-2})$.

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An electron revolves around the nucleus in a circular path with angular momentum $\vec{L}$. $A$ uniform magnetic field $\vec{B}$ is applied perpendicular to the plane of its orbit. If the electron experiences a torque $\vec{\tau}$, then

$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$

Two particles $A$ and $B$ having equal charges $+6\,C$,after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii $2\,cm$ and $3\,cm$ respectively. The ratio of mass of $A$ to that of $B$ is

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$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. If the particle moves in a straight line, the value of the magnetic induction is $(g=10 \,ms^{-2})$.

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