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Class 12 › Physics › Moving Charges and Magnetism
CBSE2023Class Class 12 · Physics5 Marks · Long✅ Verified
A point-sized object having charge 1 C and mass 1 g is projected with velocity $2\hat{i}$ m/s from point $(0, 2\text{ cm}, 0)$ in a magnetic field $\vec{B} = -0.1\hat{k}$ T in the first quadrant.
(A) What will be the shape of the path?
(B) At what point will it cross the X-axis?
(C) What will be the kinetic energy of the particle when it enters the fourth quadrant?
✅ Answer & Solution
(A) Shape of path:
The particle moves with velocity along $+x$, field along $-z$. Force $F = q(\vec{v} \times \vec{B})$ is perpendicular to both — always. Hence the path is a circle.
(B) Radius of circular path:
$$r = \frac{mv}{qB} = \frac{1 \times 10^{-3} \times 2}{1 \times 0.1} = 0.02 \text{ m} = 2 \text{ cm}$$
The particle starts at $(0, 2\text{ cm}, 0)$ with radius 2 cm. The centre of the circle is at $(2\text{ cm}, 2\text{ cm}, 0)$.
It crosses the X-axis at point $(4\text{ cm}, 0, 0)$ i.e. $(0.04\text{ m}, 0, 0)$.
(C) Kinetic energy:
Magnetic force does no work on the particle, so kinetic energy remains unchanged:
$$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 10^{-3} \times (2)^2 = 2 \times 10^{-3} \text{ J} = 2 \text{ mJ}$$
