Moving Charges and Magnetism - Result Question 36
37. When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$ it moves with an initial acceleration $3 a_0$ towards west. The electric and magnetic fields in the room are respectively
[2013]
(a) $\frac{m a_0}{e}$ west, $\frac{2 m a_0}{e v_0}$ down
(b) $\frac{m a_0}{e}$ east, $\frac{3 m a_0}{e v_0}$ up
(c) $\frac{m a_0}{e}$ east, $\frac{3 m a_0}{e v_0}$ down
(d) $\frac{m a_0}{e}$ west, $\frac{2 m a_0}{e v_0}$ up
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Answer:
Correct Answer: 37. (a)
Solution:
- (a)
When moves with an acceleration $a_0$ towards west, electric field
$E=\frac{F}{q}=\frac{m a_0}{e}$ (West)
When moves with an acceleration $3 a_0$ towards east, magnetic field
$B=\frac{2 m a_0}{e v_0}$ (downward)
