(C) The magnetic force on a moving charge is given by the formula $\vec F_m = q(\vec v \times \vec B)$.Given:Charge $q = 2.0 \times 10^{-6}\,C$Velocit

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$F_m = 1.2\,N$ along the positive $y$-direction

(C) The magnetic force on a moving charge is given by the formula $\vec F_m = q(\vec v \times \vec B)$.Given:Charge $q = 2.0 \times 10^{-6}\,C$Velocity $\vec v = 3.0 \times 10^6\,\hat i\,m/s$Magnetic field $\vec B = -0.2\,\hat k\,T$Substituting these values into the formula:$\vec F_m = (2.0 \times 10^{-6}) \times (3.0 \times 10^6\,\hat i \times -0.2\,\hat k)$$\vec F_m = (2.0 \times 3.0 \times -0.2) \times (\hat i \times \hat k)$Since $\hat i \times \hat k = -\hat j$,we have:$\vec F_m = -1.2 \times (-\hat j) = 1.2\,\hat j\,N$Therefore,the force is $1.2\,N$ along the positive $y$-direction.

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

Medium

$A$ charge of $2.0\,\mu C$ moves with a speed of $3.0 \times 10^6\,m/s$ along the positive $X$-axis. $A$ magnetic field of strength $\vec B = -0.2\,\hat k\,T$ exists in space. What is the magnetic force $(\vec F_m)$ on the charge?

A
$F_m = 1.2\,N$ along the positive $x$-direction
B
$F_m = 1.2\,N$ along the negative $x$-direction
C
$F_m = 1.2\,N$ along the positive $y$-direction
D
$F_m = 1.2\,N$ along the negative $y$-direction

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

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$A$ charge of $2.0\,\mu C$ moves with a speed of $3.0 \times 10^6\,m/s$ along the positive $X$-axis. $A$ magnetic field of strength $\vec B = -0.2\,\hat k\,T$ exists in space. What is the magnetic force $(\vec F_m)$ on the charge?

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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 . . . . . . $s$. $(\pi = 3.14)$

$A$ proton (or charged particle) moving with velocity $v$ is acted upon by an electric field $E$ and a magnetic field $B$. The proton will move undeflected if

$A$ negatively charged particle projected towards east is deflected towards north by a magnetic field. The direction of the magnetic field is

$A$ particle with $10^{-11} \, C$ of charge and $10^{-7} \, kg$ mass is moving with a velocity of $10^8 \, m/s$ along the $y$-axis. $A$ uniform static magnetic field $B = 0.5 \, T$ is acting along the $x$-direction. The force on the particle is:

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