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A proton goes undeflected in a crossed electric and magnetic field(the fields are perpendicular to each other) at a speed of $2.0 \times 10^{5} m / s$ . The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. The magnitude of the electric and the magnetic fields are:(Take the mass of the protons=1.6 $\times$ $10^{- 27}$ kg)

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$10^{4} \frac{N}{C}, 0.05 T$

$10^{5} \frac{N}{C}, 0.05 T$

$10^{- 4} \frac{N}{C}, 0.05 T$

$10^{4} \frac{N}{C}, 0.04 T$

answer is A.

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Detailed Solution

$B q v = q E \Rightarrow E = B v$

or $E = 2.0 \times 10^{5} B$ …(i)

$r = \frac{m v}{B q}$

or $4 \times 10^{- 2} = \frac{1.6 \times 10^{- 27} \times 2.0 \times 10^{5}}{(B) (1.6 \times 10^{- 19})}$ …(ii)

Solving Eqs. (i) and (ii), we can find $E$ and $B$ .

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