Question

The magnetic force acting on a charged particle carrying a charge $3 \mu C$ in a magnetic field of $5$ T acting in the $y$-direction, when the particle velocity is \[ (\hat{i} + \hat{j}) \times 10^5 { ms}^{-1} \] is

• A. $0.5$ N in $+x$ direction
• B. $0.2$ N in $+y$ direction
• C. $2$ N in $-x$ direction
• D. $1.5$ N in $-z$ direction
• E. $1.5$ N in $+z$ direction
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Hint:

Use the determinant method to compute vector cross products efficiently.

Answer 1

Answer: (E)

Step 1: The magnetic force on a moving charge is given by: \[ \vec{F} = q (\vec{v} \times \vec{B}) \] 
Step 2: Given: \[ q = 3 \times 10^{-6} C, \quad \vec{v} = (10^5 \hat{i} + 10^5 \hat{j}) { m/s}, \quad \vec{B} = 5 \hat{j} { T} \] 
Step 3: Compute the cross product:

Step 4: Expanding the determinant: \[ \vec{v} \times \vec{B} = (10^5 \times 0 - 10^5 \times 0) \hat{i} - (10^5 \times 0 - 5 \times 10^5) \hat{j} + (10^5 \times 5 - 10^5 \times 0) \hat{k} \] \[ = 0 \hat{i} + 5 \times 10^5 \hat{j} + 5 \times 10^5 \hat{k} \] \[ = 5 \times 10^5 \hat{k} \] 
Step 5: Compute force: \[ \vec{F} = (3 \times 10^{-6}) (5 \times 10^5 \hat{k}) \] \[ = 1.5 \hat{k} { N} \] 
Step 6: Since $\hat{k}$ represents the $+z$ direction, the force is $1.5$ N in the $+z$ direction. 
Step 7: Therefore, the correct answer is (E).