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Magnetics question

2024 · 31 Jan · Shift 1 · Q82
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  5. /2024 · 31 Jan · Shift 1 · Q82

Magnetics question

2024 · 31 Jan · Shift 1 · Q82

JEE MainPhysicsMagneticsNumerical+4 / −1
An electron moves through a uniform magnetic field B⃗=B0i^+2B0j^T\vec{B}=B_0 \hat{i}+2 B_0 \hat{j} TB=B0​i^+2B0​j^​T. At a particular instant of time, the velocity of electron is u⃗=3i^+5j^ m/s\vec{u}=3 \hat{i}+5 \hat{j} \mathrm{~m} / \mathrm{s}u=3i^+5j^​ m/s. If the magnetic force acting on electron is F⃗=5ek^N\vec{F}=5 e \hat{k} NF=5ek^N, where eee is the charge of electron, then the value of B0B_0B0​ is ‾\underline{\hspace{2cm}}​TTT.
Numerical answer
View written solutionFree

Correct answer: -5

  1. Given data

    Magnetic field: B⃗=B0i^+2B0j^\vec{B}=B_0\hat{i}+2B_0\hat{j}B=B0​i^+2B0​j^​

    Velocity of electron: u⃗=3i^+5j^ m/s\vec{u}=3\hat{i}+5\hat{j}\ \text{m/s}u=3i^+5j^​ m/s

    Magnetic force on a charge is: F⃗=q(u⃗×B⃗)\vec{F}=q(\vec{u}\times\vec{B})F=q(u×B)

    For an electron, q=−eq=-eq=−e

    Given force: F⃗=5ek^ N\vec{F}=5e\hat{k}\ \text{N}F=5ek^ N

  2. Compute u⃗×B⃗\vec{u}\times\vec{B}u×B

    \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 5 & 0\\ B_0 & 2B_0 & 0 \end{vmatrix}$$ Expanding, $$\vec{u}\times\vec{B}=\hat{k}(3\cdot 2B_0-5\cdot B_0)$$ $$\vec{u}\times\vec{B}=\hat{k}(6B_0-5B_0)=B_0\hat{k}$$

  3. Now calculate the force on electron

    F⃗=q(u⃗×B⃗)\vec{F}=q(\vec{u}\times\vec{B})F=q(u×B) F⃗=(−e)(B0k^)=−eB0k^\vec{F}=(-e)(B_0\hat{k})=-eB_0\hat{k}F=(−e)(B0​k^)=−eB0​k^

  4. Compare with the given force

    Given, F⃗=5ek^\vec{F}=5e\hat{k}F=5ek^

    So, −eB0k^=5ek^-eB_0\hat{k}=5e\hat{k}−eB0​k^=5ek^

    Cancelling ek^e\hat{k}ek^, −B0=5-B_0=5−B0​=5 B0=−5B_0=-5B0​=−5

  5. Final value

    −5\boxed{-5}−5​

    Since this is an integer-type question, the required answer is: −5\boxed{-5}−5​

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