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NEET 2013
Exam
NEET
Year
2013
Subject
Physics
Chapter
Magnetic Effects of Current

NEET 2013 · Physics · Magnetic Effects of Current

When a proton is released from rest in a room, it starts with an...

When a proton is released from rest in a room, it starts with an initial acceleration a0a_0 towards west. When it is projected towards north with a speed v0v_0 it moves with an initial acceleration 3a03 a_0 towards west. The electric and magnetic fields in the room are

Options

  1. Ama0e\frac{m a_0}{e} west, 2ma0ev0\frac{2 m a_0}{e v_0} up
  2. Bma0e\frac{m a_0}{e} west, 2ma0ev0\frac{2 m a_0}{e v_0} down
  3. Cma0e\frac{m a_0}{e} east, 3ma0ev0\frac{3 m a_0}{e v_0} up
  4. Dma0e\frac{m a_0}{e} east 3ma0ev0\frac{3 m a_0}{e v_0} down

Correct answer

B. ma0e\frac{m a_0}{e} west, 2ma0ev0\frac{2 m a_0}{e v_0} down

Step-by-step solution

Initial acceleration, a0=∂Ema_0=\frac{\partial E}{m}⇒E=a0me∴ev0B+eEm=3a0 or ev0B+eE=3a0m∴ev0B=3ma0−eE⇒=3ma0−ma0[ fromeq. (i)] ⇒ev0B=2ma0∴B=2ma0ev0\begin{aligned} & \Rightarrow \quad E=\frac{a_0 m}{e} \therefore \frac{e v_0 B+e E}{m}=3 a_0 \\ & \text{ or } \quad e v_0 B+e E=3 a_0 m \\ & \therefore \quad e v_0 B=3 m a_0-e E \\ & \Rightarrow \quad=3 m a_0-m a_0 \quad[\text{ fromeq. (i)] } \\ & \Rightarrow \quad e v_0 B=2 m a_0 \\ & \therefore \quad B=\frac{2 m a_0}{e v_0} \\ & \end{aligned}$$
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