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MHT CET 202625 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
25 April
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

MHT CET 2026 · 25 April · Morning Shift · Physics · Magnetic Effects of Current

A proton moves at a constant speed v along the x-axis through a...

A proton moves at a constant speed vv along the x-axis through a crossed electric and magnetic field. The magnetic field is B⃗=B0k^\vec{B} = B_0\hat{k} T. The electric field E⃗\vec{E} in V/m is

Options

  1. A−vB0j^-vB_0\hat{j}
  2. BvB0j^vB_0\hat{j}
  3. CvB0k^vB_0\hat{k}
  4. DvB0i^vB_0\hat{i}

Correct answer

B. vB0j^vB_0\hat{j}

Step-by-step solution

Velocity of the proton, v⃗=vi^\vec{v} = v\hat{i}

Magnetic field, B⃗=B0k^\vec{B} = B_0\hat{k}

For the proton to move at a constant velocity, the net force acting on it must be zero.

Lorentz force, F⃗=q(E⃗+v⃗×B⃗)=0\vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) = 0

E⃗=−(v⃗×B⃗)\vec{E} = -(\vec{v} \times \vec{B})

E⃗=−(vi^×B0k^)\vec{E} = -(v\hat{i} \times B_0\hat{k})

E⃗=−vB0(i^×k^)\vec{E} = -vB_0(\hat{i} \times \hat{k})

Since i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j}

E⃗=−vB0(−j^)=vB0j^\vec{E} = -vB_0(-\hat{j}) = vB_0\hat{j}

Answer: vB0j^vB_0\hat{j}
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