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=?

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answer is 0012.80.

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Detailed Solution

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