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

V→×B→+QE→=0⇒E→=QV→×B→Therefore E→ and B→ must be mutually perpendicular. i.e. E→. B→ = 0 ⇒3i^-3j^.ai^-5j^=0⇒3a-15=0⇒a=5∴B=52+52T=52T

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

V→×B→+QE→=0⇒E→=QV→×B→Therefore E→ and B→ must be mutually perpendicular. i.e. E→. B→ = 0 ⇒3i^-3j^.ai^-5j^=0⇒3a-15=0⇒a=5∴B=52+52T=52T

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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

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$\frac{7}{2} T$

$\frac{9}{2} T$

$4 T$

$5 \sqrt{2} T$

answer is A.

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$\vec{V} \times \vec{B} + Q \vec{E} = 0 \Rightarrow \vec{E} = Q \vec{V} \times \vec{B}$

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

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