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  4. A charge having elm equal to 108 C/kg and with velocity 3 × 10 5 m/s enters into a uniform magnetic field B = 0.3 tesla at an angle 30° with direction of field. The radius of curvature will be

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Q. A charge having elm equal to $ 10^8 C/kg $ and with velocity $ 3 \times 10^ 5 m/s $ enters into a uniform magnetic field B = 0.3 tesla at an angle 30$^{\circ}$ with direction of field. The radius of curvature will be

AIPMTAIPMT 1999Moving Charges and Magnetism

Solution:

$r =\frac{ mV _{\perp}}{ qB }$
$r =\left|\frac{ m }{ q }\right|\left|\frac{3 \times 10^{5} \times \sin 30^{\circ}}{0.3}\right|$
$r =\frac{3 \times 105}{10^{8} \times 0.3 \times 2}=0.5 \times 10^{-2} m =0.5 cm$