Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

107 of 276

Question 107 of 276

The charged particle moving in a uniform magnetic field of (3i^+2j^)(3\hat{i} + 2\hat{j}) T has an acceleration (4i^−x2j^)\left(4\hat{i} - \dfrac{x}{2}\hat{j}\right) m/s2^2. The value of xx is __________.

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Answer: 12

The magnetic force on a moving charged particle is given by F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B}).

Since the acceleration a⃗\vec{a} is in the direction of the magnetic force, it must be perpendicular to the magnetic field B⃗\vec{B}.

Therefore, the dot product of acceleration and magnetic field is zero:

a⃗⋅B⃗=0\vec{a} \cdot \vec{B} = 0

Substituting the given vectors:

(4i^−x2j^)⋅(3i^+2j^)=0\left(4\hat{i} - \dfrac{x}{2}\hat{j}\right) \cdot (3\hat{i} + 2\hat{j}) = 0

(4)(3)+(−x2)(2)=0(4)(3) + \left(-\dfrac{x}{2}\right)(2) = 0

12−x=012 - x = 0

x=12x = 12

Answer: 1212

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