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Magnetics question

2022 · 26 Jul · Shift 1 · Q51
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  5. /2022 · 26 Jul · Shift 1 · Q51

Magnetics question

2022 · 26 Jul · Shift 1 · Q51

JEE MainPhysicsMagneticsMCQ+4 / −1
A charge particle is moving in a uniform magnetic field (2i^+3j^) T(2 \hat{i}+3 \hat{j}) \,\mathrm{T}(2i^+3j^​)T. If it has an acceleration of (αi^−4j^) m/s2(\alpha \hat{i}-4 \hat{j})\, \mathrm{m} / \mathrm{s}^{2}(αi^−4j^​)m/s2, then the value of α\alphaα will be :
  1. A
    3
  2. B
    6
  3. C
    12
  4. D
    2
View written solutionFree

Correct answer: B

  1. Use the magnetic force relation

For a charged particle moving in a magnetic field,

F⃗=q(v⃗×B⃗)\vec F = q(\vec v \times \vec B)F=q(v×B)

So the acceleration is

a⃗=F⃗m=qm(v⃗×B⃗)\vec a = \frac{\vec F}{m} = \frac{q}{m}(\vec v \times \vec B)a=mF​=mq​(v×B)

Hence, a⃗\vec aa must be perpendicular to B⃗\vec BB.

So,

a⃗⋅B⃗=0\vec a \cdot \vec B = 0a⋅B=0
  1. Write the given vectors

Magnetic field:

B⃗=2i^+3j^\vec B = 2\hat i + 3\hat jB=2i^+3j^​

Acceleration:

a⃗=αi^−4j^\vec a = \alpha \hat i - 4\hat ja=αi^−4j^​
  1. Apply perpendicular condition
a⃗⋅B⃗=(αi^−4j^)⋅(2i^+3j^)=0\vec a \cdot \vec B = (\alpha \hat i - 4\hat j) \cdot (2\hat i + 3\hat j) = 0a⋅B=(αi^−4j^​)⋅(2i^+3j^​)=0

Compute dot product:

2α+(−4)(3)=02\alpha + (-4)(3) = 02α+(−4)(3)=0 2α−12=02\alpha - 12 = 02α−12=0 2α=122\alpha = 122α=12 α=6\alpha = 6α=6
  1. Check options

The correct option is:

6\boxed{6}6​

which is Option B.

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