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Standard XII

Physics

Circular Kinematics

A charge havi...

Question

A charge having q/m equal to $10^{8} c / k g$ and with velocity $3 \times 10^{5} m / s$ enters into a uniform magnetic field B = 0.3 tesla at an angle $30^{0}$ with direction of field. Then radius of curvature will be

0.01 cm

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0.5 cm

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1 cm

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2 cm

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Solution

The correct option is A 0.5 cm
$r = \frac{m V_{⊥}}{q B}$
$r = ∣ ∣ ∣ \frac{m}{q} ∣ ∣ ∣ ∣ ∣ ∣ \frac{3 \times 10^{5} \times s i n 30^{o}}{0.3} ∣ ∣ ∣$
$r = \frac{3 \times 105}{10^{8} \times 0.3 \times 2} = 0.5 \times 10^{- 2} m = 0.5 c m$

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A charge having q/m equal to 108c/kg and with velocity 3×105m/senters into a uniform magnetic field B = 0.3 tesla at an angle 300 with direction of field. Then radius of curvature will be

The correct option is A 0.5 cmr=mV⊥qBr=∣∣∣mq∣∣∣∣∣∣3×105×sin30o0.3∣∣∣r=3×105108×0.3×2=0.5×10−2m=0.5cm

A charge having q/m equal to $10^{8} c / k g$ and with velocity $3 \times 10^{5} m / s$ enters into a uniform magnetic field B = 0.3 tesla at an angle $30^{0}$ with direction of field. Then radius of curvature will be

The correct option is A 0.5 cm
$r = \frac{m V_{⊥}}{q B}$
$r = ∣ ∣ ∣ \frac{m}{q} ∣ ∣ ∣ ∣ ∣ ∣ \frac{3 \times 10^{5} \times s i n 30^{o}}{0.3} ∣ ∣ ∣$
$r = \frac{3 \times 105}{10^{8} \times 0.3 \times 2} = 0.5 \times 10^{- 2} m = 0.5 c m$