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A charge having $\dfrac{q}{m}$ 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. Then radius of curvature will be :
A
0.01 cm
B
0.5 cm
C
1 cm
D
2 cm
Detailed Solution
Only the velocity component perpendicular to the field causes circular motion: $v_\perp = v\sin 30^\circ$.
[add image: velocity V at $30^\circ$ to B, with components V cos30 along B and V sin30 perpendicular to B]
$r = \dfrac{mv_\perp}{qB} = \dfrac{m}{q} \cdot \dfrac{v\sin 30^\circ}{B}$
$r = \dfrac{3 \times 10^5 \times 0.5}{10^8 \times 0.3} = \dfrac{3 \times 10^5}{10^8 \times 0.3 \times 2}$
$r = 0.5 \times 10^{-2}$ m $= 0.5$ cm
(The particle moves along a helix of this radius.)
[add image: velocity V at $30^\circ$ to B, with components V cos30 along B and V sin30 perpendicular to B]
$r = \dfrac{mv_\perp}{qB} = \dfrac{m}{q} \cdot \dfrac{v\sin 30^\circ}{B}$
$r = \dfrac{3 \times 10^5 \times 0.5}{10^8 \times 0.3} = \dfrac{3 \times 10^5}{10^8 \times 0.3 \times 2}$
$r = 0.5 \times 10^{-2}$ m $= 0.5$ cm
(The particle moves along a helix of this radius.)