Magnetic Effects & Magnetism

JEE Main level, single correct, medium

134 of 276

Question 134 of 276

A charge particle is moving in a uniform magnetic field (2i^+3j^)T\left(2\hat{i}+3\hat{j}\right)T. If it has an acceleration of (αi^−4j^)ms−2\left(\alpha \hat{i}-4\hat{j}\right)m{s}^{-2}, then the value of α\alpha will be

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Answer: (B) 66

The magnetic force on a moving charged particle is given by F⃗=q(v⃗×B⃗)\vec{F}=q\left(\vec{v}\times \vec{B}\right). Therefore, we know that the magnetic force is perpendicular to the magnetic field. Therefore, the resulting acceleration will also be perpendicular to the magnetic field. As the scalar product of two mutually perpendicular vectors is zero,
⇒a⃗⋅B⃗=0⇒(αi^−4j^)⋅(2i^+3j^)=0\Rightarrow \vec{a}\cdot \vec{B}=0 \Rightarrow \left(\alpha \hat{i}-4\hat{j}\right)\cdot \left(2\hat{i}+3\hat{j}\right)=0
⇒2α−12=0⇒α=6\Rightarrow 2\alpha -12=0 \Rightarrow \alpha =6

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