A proton beam moves through a region of space where there exists a uniform magnitude $$4.0$$ T along $$z-axis$$. The protons have a velocity of $$4$$×$$105m/s$$ in the $$x-z$$ plane at an angle $$300$$ to the positive $$z-axis$$. Then the force acting on the proton is −x×$$10$$−$$14j^$$ N, find $$x=$$?

∴ F→$$=qv$$→×B→$$=1.6$$×$$10$$−$$194$$×$$105)$$×[$$(sin30oi^+cos30ok^$$×$$4k^$$] ⇒F→$$=$$−$$12.8$$×$$10$$−$$14j^$$ ∴ $$x=12.8$$

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A proton beam moves through a region of space where there exists a uniform magnitude 4.0 T along z-axis. The protons have a velocity of $4 \times 10^{5} m / s$ in the x-z plane at an angle $30^{0}$ to the positive z-axis. Then the force acting on the proton is $- x \times 10^{- 14} \hat{j} N$ , find $x = ?$

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Solution

$∴ \vec{F} = q (\vec{v} \times \vec{B}) = 1.6 \times 10^{- 19} (4 \times 10^{5}) \times [(\sin 30^{o} \hat{i} + \cos 30^{o} \hat{k}) \times 4 \hat{k}] \Rightarrow \vec{F} = - 12.8 \times 10^{- 14} \hat{j} ∴ x = 12.8$

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Magnetic force

Remember concepts with our Masterclasses.

A proton beam moves through a region of space where there exists a uniform magnitude 4.0 T along z-axis. The protons have a velocity of 4×105m/s in the x-z plane at an angle 300 to the positive z-axis. Then the force acting on the proton is −x×10−14j^ N, find x=?

∴ F→=qv→×B→=1.6×10−194×105)×[(sin30oi^+cos30ok^×4k^] ⇒F→=−12.8×10−14j^ ∴ x=12.8

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$∴ \vec{F} = q (\vec{v} \times \vec{B}) = 1.6 \times 10^{- 19} (4 \times 10^{5}) \times [(\sin 30^{o} \hat{i} + \cos 30^{o} \hat{k}) \times 4 \hat{k}] \Rightarrow \vec{F} = - 12.8 \times 10^{- 14} \hat{j} ∴ x = 12.8$