A charge particle with charge q enters a region of constant E and B with velocity V perpendicular to both E and B, and comes out without any charge in magnitude or direction of V, then establish a relation between V, E and B.

Solution

A charged particle goes undeflected when,
qE = qvB

$\Rightarrow$ $\Rightarrow$ v = E/B

The force due to electric field and magnetic field are acting on the charged particle and will oppose each other if $\overset{⇀}{v} i s a l o n g \overset{⇀}{E} \times \overset{⇀}{B}$ .
Now, v = $\frac{E B}{B^{2}} = \frac{E B \sin 90^{o}}{B^{2}} = \frac{|\overset{⇀}{E} \times \overset{⇀}{B}|}{B^{2}}$
Therefore,

$\overset{⇀}{v} = \frac{|\overset{⇀}{E} \times \overset{⇀}{B}|}{B^{2}}$
which is the required relation . $Error: timeout waiting for response.$

Some More Questions From Moving Charges And Magnetism Chapter

Answer the following questions:
A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

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Moving Charges And Magnetism

A charge particle with charge q enters a region of constant E and B with velocity V perpendicular to both E and B, and comes out without any charge in magnitude or direction of V, then establish a relation between V, E and B.

Some More Questions From Moving Charges And Magnetism Chapter

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