(C) When a charged particle passes undeflected through mutually perpendicular electric and magnetic fields,the electric force $(F_e)$ must balance the

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(C) When a charged particle passes undeflected through mutually perpendicular electric and magnetic fields,the electric force $(F_e)$ must balance the magnetic force $(F_m)$.$F_e = F_m$$qE = qvB \sin \theta$Since the fields are mutually perpendicular,$\theta = 90^{\circ}$,so $\sin 90^{\circ} = 1$.Thus,$E = vB$.Given:Velocity $v = 2 \times 10^{3} \ ms^{-1}$Magnetic field $B = 1.5 \ T$$E = (2 \times 10^{3}) \times 1.5 = 3 \times 10^{3} \ V/m$ (or $NC^{-1}$).Therefore,the magnitude of the electric field is $3 \times 10^{3} \ NC^{-1}$.

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

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$A$ charged particle with a velocity $2 \times 10^{3} \ ms^{-1}$ passes undeflected through electric and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5 \ T$. The magnitude of the electric field will be:

KCET 2013 All KCET

A
$1.5 \times 10^{3} \ NC^{-1}$
B
$2 \times 10^{3} \ NC^{-1}$
C
$3 \times 10^{3} \ NC^{-1}$
D
$1.33 \times 10^{3} \ NC^{-1}$

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

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$A$ charged particle with a velocity $2 \times 10^{3} \ ms^{-1}$ passes undeflected through electric and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5 \ T$. The magnitude of the electric field will be:

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$A$ uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5 \ Wb/m^2$. If a proton having mass $m = 1.7 \times 10^{-27} \ kg$ and charge $q = 1.6 \times 10^{-19} \ C$ moves in this field vertically downwards with energy $5 \ MeV$,then the force acting on it will be

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