Magnetic Effects & Magnetism

JEE Main level, single correct, medium

108 of 276

Question 108 of 276

A particle having charge 10−910^{-9} C moving in xx-yy plane in fields of 0.4j^0.4\hat{j} N/C and 4×10−3k^4 \times 10^{-3} \hat{k} T experiences a force of (4i^+2j^)×10−10(4\hat{i} + 2\hat{j}) \times 10^{-10} N. The velocity of the particle at that instant is _______ m/s.

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Answer: (A) 50i^+100j^50\hat{i} + 100\hat{j}

The Lorentz force experienced by a charged particle is given by F⃗=q(E⃗+v⃗×B⃗)\vec{F} = q(\vec{E} + \vec{v} \times \vec{B}).

Given that the particle moves in the xx-yy plane, its velocity can be written as v⃗=vxi^+vyj^\vec{v} = v_x \hat{i} + v_y \hat{j}.

Substituting the given values into the force equation:

F⃗=10−9[0.4j^+(vxi^+vyj^)×(4×10−3k^)]\vec{F} = 10^{-9} [0.4\hat{j} + (v_x \hat{i} + v_y \hat{j}) \times (4 \times 10^{-3} \hat{k})]

Using the cross product rules i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j} and j^×k^=i^\hat{j} \times \hat{k} = \hat{i}:

F⃗=10−9[0.4j^−4×10−3vxj^+4×10−3vyi^]\vec{F} = 10^{-9} [0.4\hat{j} - 4 \times 10^{-3} v_x \hat{j} + 4 \times 10^{-3} v_y \hat{i}]

F⃗=10−9[4×10−3vyi^+(0.4−4×10−3vx)j^]\vec{F} = 10^{-9} [4 \times 10^{-3} v_y \hat{i} + (0.4 - 4 \times 10^{-3} v_x) \hat{j}]

The given force is F⃗=(4i^+2j^)×10−10\vec{F} = (4\hat{i} + 2\hat{j}) \times 10^{-10} N, which can be rewritten as:

F⃗=10−9(0.4i^+0.2j^)\vec{F} = 10^{-9} (0.4\hat{i} + 0.2\hat{j}) N

Comparing the coefficients of i^\hat{i} and j^\hat{j} from both expressions:

For the i^\hat{i} component:
4×10−3vy=0.44 \times 10^{-3} v_y = 0.4
vy=0.44×10−3=100v_y = \dfrac{0.4}{4 \times 10^{-3}} = 100 m/s

For the j^\hat{j} component:
0.4−4×10−3vx=0.20.4 - 4 \times 10^{-3} v_x = 0.2
4×10−3vx=0.24 \times 10^{-3} v_x = 0.2
vx=0.24×10−3=50v_x = \dfrac{0.2}{4 \times 10^{-3}} = 50 m/s

Therefore, the velocity of the particle is v⃗=50i^+100j^\vec{v} = 50\hat{i} + 100\hat{j} m/s.

Answer: 50i^+100j^50\hat{i} + 100\hat{j}

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