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QPages 230-234: Q3: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q3 Solution: A magnetic field of exerts a force of on a particle with a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently ignore the negative sign in the $y$-component of the cross product ($\vec{v} \times \vec{B}$) when $\vec{B}$ is in the $\hat{k}$ direction, leading to an incorrect positive value for $v_x$. Always expand the vector product explicitly to capture the sign change: $\vec{v} \times \hat{k} = -v_x \hat{j} + v_y \hat{i}$.

A magnetic field of 4.0×10−3k^) T4.0 \times 10^{-3} \hat{k}) \text{ T} exerts a force of 4.0i^+3.0j^)×10−10 N4.0 \hat{i} + 3.0 \hat{j}) \times 10^{-10} \text{ N} on a particle with a charge of 1.0×10−9 C1.0 \times 10^{-9} \text{ C} and going in the x−yx - y plane. Find the velocity of the particle.

Step-by-Step Verified Solution

**Answer:** v⃗=(−75i^+100j^) m/s\vec{v} = (-75\hat{i} + 100\hat{j}) \text{ m/s}. **Step-by-Step Explanation:** Given: B⃗=4×10−3k^ T\vec{B} = 4 \times 10^{-3} \hat{k} \text{ T}, F⃗=(4i^+3j^)×10−10 N\vec{F} = (4\hat{i} + 3\hat{j}) \times 10^{-10} \text{ N}, q=1×10−9 Cq = 1 \times 10^{-9} \text{ C}. From F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B}): Fx=q(vyBz)=1×10−9(vy)(4×10−3)=4×10−10⇒vy=100 m/sF_x = q(v_y B_z) = 1 \times 10^{-9} (v_y)(4 \times 10^{-3}) = 4 \times 10^{-10} \Rightarrow v_y = 100 \text{ m/s}. Fy=−q(vxBz)=−1×10−9(vx)(4×10−3)=3×10−10⇒vx=−75 m/sF_y = -q(v_x B_z) = -1 \times 10^{-9} (v_x)(4 \times 10^{-3}) = 3 \times 10^{-10} \Rightarrow v_x = -75 \text{ m/s}.

All Questions in this Chapter

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