The force on a charged particle due to electric and magnetic fields is given by F-q E-q v-B- Suppose vecE is along the X -axis and B- along the Y-axis- In what direction and with what minim-um speed v should a positively charged particle be sent so that the net force on it is zero-

Given, F⃗ = qE⃗ + q(v⃗×B⃗) = 0 ⇒ E = (v⃗×B⃗) So, the direction of v × B should be opposite to the direction of E⃗. Hence v⃗ should be in the +ve yz plane. A ...

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Standard XII

Physics

Coulomb's Law - Mathematical Form

The force on ...

Question

The force on a charged particle due to electric and magnetic fields is given by $\to F = q \to E + q \to v \times \to B$ . Suppose \vec{E} is along the X-axis and $\to B$ along the Y-axis. In what direction and with what minim.um speed v should a positively charged particle be sent so that the net force on it is zero?

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Solution

Given, $\to F = q \to E + q (\to v \times \to B) = 0$

$\Rightarrow E = - (\to v \times \to B)$

So, the direction of v $\times$ B should be opposite to the direction of $\to E$ . Hence $\to v$ should be in the +ve yz - plane.

Again, e = vB sin $θ$

$\Rightarrow v = \frac{E}{B} s i n θ$

For v to be minimum,

$θ = 90^{0}$ and so $v_{m i n} = \frac{E}{B}$

So, the particle must be projected at minimum speed of $\frac{E}{B}$ along '-' ve z -axis

$(θ = 90^{0})$ as shown in the figure so that the force is zero.

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The force on a charged particle due to electric and magnetic fields is given by →F=q→E+q→v×→B. Suppose \vec{E} is along the X-axis and →B along the Y-axis. In what direction and with what minim.um speed v should a positively charged particle be sent so that the net force on it is zero?