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Q.

A proton beam passes without deviation through a region of space where there are transverse mutually perpendicular electric and magnetic fields with $E = 120 \frac{V}{m}$ and $B = 50 m T$ .Then the beam strikes a grounded target. Find the force imparted by the force on the target if the beam current is equal to $I = 0.80 m A$ .

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a

$2 \times 10^{- 5} N$

b

$2 \times 10^{5} N$

c

$3 \times 10^{- 5} N$

d

$5 \times 10^{- 5} N$

answer is A.

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Detailed Solution

$F_{e} = F_{m}; e E = e B v ∴ v = \frac{E}{B} = \frac{120 \times 10^{3}}{50 \times 10^{- 3}} = 2.4 \times 10^{6} m / s$

Let $n$ be the number of protons striking per second. Then,

$n e = 0.8 \times 10^{- 3} n = \frac{0.8 \times 10^{- 3}}{1.6 \times 10^{- 19}} = 5 \times 10^{15} m / s$

Force imparted= Rate of change of momentum

$= n m v = 5 \times 10^{15} \times 1.67 \times 10^{- 27} \times 2.4 \times 10^{6} = 2.0 \times 10^{- 5} N$

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A proton beam passes without deviation through a region of space where there are transverse mutually perpendicular electric and magnetic fields with E=120Vm and B=50 mT.Then the beam strikes a grounded target. Find the force imparted by the force on the target if the beam current is equal to I=0.80 mA.