First slide

Force on a charged particle moving in a magnetic field

Question

A charged particle carrying a charge 2µC is fired in a gravity free space where mutually perpendicular electric and magnetic field exist. If $\vec{E} = (3 \hat{i} - 3 \hat{j}) v / m$ and $\vec{B} = (a \hat{i} + 5 \hat{j}) Tesla$ and the particle passes undeviated, then magnitude of magnetic field is

Moderate

A

$5 \sqrt{2} T$
B

$\frac{7}{2} T$
C

$4 T$
D

$\frac{9}{2} T$

Solution

$\vec{V} \times \vec{B} + Q \vec{E} = 0 \Rightarrow \vec{E} = Q \vec{V} \times \vec{B}$
Therefore $\vec{E}$ and $\vec{B}$ must be mutually perpendicular. i.e. $\vec{E} . \vec{B} = 0 \Rightarrow (3 \hat{i} - 3 \hat{j}) . (a \hat{i} - 5 \hat{j}) = 0 \Rightarrow 3 a - 15 = 0 \Rightarrow a = 5$
$∴ B = \sqrt{5^{2} + 5^{2}} T = 5 \sqrt{2} T$

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