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(D) Given: Charge $q = 4.0 \mu C = 4.0 	imes 10^{-6} \ C$.Velocity $\vec{v} = 4.0 	imes 10^6 \hat{j} \ m/s$.Magnetic field $\vec{B} = 2 \hat{k} \ T$

Vedclass · Accepted Answer

$32$

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(D) Given: Charge $q = 4.0 \mu C = 4.0 	imes 10^{-6} \ C$.Velocity $\vec{v} = 4.0 	imes 10^6 \hat{j} \ m/s$.Magnetic field $\vec{B} = 2 \hat{k} \ T$.The magnetic force $\vec{F}$ on a moving charge is given by the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.Substituting the values: $\vec{F} = (4.0 	imes 10^{-6} \ C) 	imes (4.0 	imes 10^6 \hat{j} \ m/s 	imes 2 \hat{k} \ T)$.Using the cross product rule $\hat{j} 	imes \hat{k} = \hat{i}$,we get: $\vec{F} = (4.0 	imes 10^{-6}) 	imes (8.0 	imes 10^6) \hat{i} \ N$.$\vec{F} = 32 \hat{i} \ N$.Comparing this with $\vec{F} = x \hat{i} \ N$,we find $x = 32$.

$A$ charge of $4.0 \mu C$ is moving with a velocity of $4.0 \times 10^6 \ m/s$ along the positive $y$-axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k}) \ T$. The force acting on the charge is $x \hat{i} \ N$. The value of $x$ is . . . . . . .

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