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A particle having a charge 20mC and mass 20mg moves along a circle of radius 5.0 cm under the action of a magnetic field B = 1.0T. When the particle is at a point P, a uniform electric field is switched on and it is found that the particle continues on the tangent through P with a uniform velocity. The electric field is.

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25 V/m

100V/m

150V/m

50V/m

answer is C.

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Detailed Solution

q = 20 mC = 20 \times 10^{- 3} C

$m = 20 mg = 20 \times 10^{- 6} kg$

$B = 1.0 T$

since particle is moving along straight line when E and B are simultaneously acted

$F_{e} = F_{m}$

$qE = Bqv$

$\Rightarrow v = \frac{E}{B}$

$E = Bv ∵ v = \frac{Bqr}{m}$

Now speed $v = \frac{1 \times 20 \times 10^{- 3} \times 5 \times 10^{- 2}}{20 \times 10^{- 6}}$

$v = 50 m/s$

Now $E = v \times B$

$= 50 \times 1 = 50 volt/m$

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