The magnetic force acting on a charged particle of charge $-2 \mu C$ in a magnetic field of $2 T$ acting in y-

Question

The magnetic force acting on a charged particle of charge $-2 \mu C$ in a magnetic field of $2 T$ acting in y-direction, when the particle velocity is $(2\hat{i} + 3\hat{j}) 	imes 10^6 	ext{ ms}^{-1}$ is:

Accepted Answer

1. **Formula:** The magnetic force $\vec{F}$ exerted on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force equation: $\vec{F} = q(\vec{v} 	imes \vec{B})$ .
2. **Given Values:**
   - Charge $q = -2 \mu	ext{C} = -2 	imes 10^{-6} 	ext{ C}$
   - Velocity $\vec{v} = (2\hat{i} + 3\hat{j}) 	imes 10^6 	ext{ ms}^{-1}$
   - Magnetic Field $\vec{B} = 2\hat{j} 	ext{ T}$
3. **Calculation:**
   - First, calculate the cross product $\vec{v} 	imes \vec{B}$:
     $$ \vec{v} 	imes \vec{B} = [(2\hat{i} + 3\hat{j}) 	imes 10^6] 	imes (2\hat{j}) $$
     $$ = 10^6 [ (2\hat{i} 	imes 2\hat{j}) + (3\hat{j} 	imes 2\hat{j}) ] $$
   - Using unit vector properties ($\hat{i} 	imes \hat{j} = \hat{k}$ and $\hat{j} 	imes \hat{j} = 0$):
     $$ \vec{v} 	imes \vec{B} = 10^6 (4\hat{k} + 0) = 4 	imes 10^6 \hat{k} $$
   - Now, multiply by the charge $q$:
     $$ \vec{F} = (-2 	imes 10^{-6}) 	imes (4 	imes 10^6 \hat{k}) $$
     $$ \vec{F} = -8 \hat{k} 	ext{ N} $$
4. **Conclusion:** The magnitude of the force is $8 	ext{ N}$, and the negative sign with $\hat{k}$ indicates it acts in the **negative z-direction** .

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