CBSE Class 12 Physics Chapter 4 Moving Charges and Magnetism NCERT Solutions

Q1. Two charged particles traverse identical helical paths in a uniform magnetic field $$\backslash mathbf\{B\}$ = $\backslash mathbf\{B\}$_0 $\backslash hat\{\backslash mathbf\{k\}\}$$ but in opposite senses. Which of the following conditions must be satisfied?

• A They have equal z-components of momenta.
• B They must have equal charges.
• C They necessarily represent a particle, anti-particle pair.
• D The charge to mass ratio satisfies $$\backslash left$($\backslash frac\{e\}\{m\}$$\backslash right$)_1 + $\backslash left$($\backslash frac\{e\}\{m\}$$\backslash right$)_2 = 0$.

Explanation: For identical helical paths in opposite senses, the pitch must be the same, and the charge-to-mass ratios must be equal in magnitude but opposite in sign. This leads to the sum of their charge-to-mass ratios being zero.

Q2. According to the Biot-Savart law, the magnetic field $$\backslash mathbf\{B\}$$ produced by a moving electron with velocity $$\backslash nu$$ is such that:

• A B is perpendicular to v.
• B B is parallel to v.
• C B obeys an inverse cube law.
• D B is along the line joining the electron and the point of observation.

Explanation: The Biot-Savart law states that the magnetic field element $d$\backslash mathbf\{B\}$$ is proportional to $I d$\backslash mathbf\{l\}$ $\backslash times$ $\backslash mathbf\{r\}$/$r^3$$. For an electron, $I d$\backslash mathbf\{l\}$$ is in the direction opposite to its velocity $$\backslash mathbf\{v\}$$. The cross product $d$\backslash mathbf\{l\}$ $\backslash times$ $\backslash mathbf\{r\}$$ results in a magnetic field $$\backslash mathbf\{B\}$$ that is perpendicular to both $d$\backslash mathbf\{l\}$$ (and thus to $$\backslash mathbf\{v\}$$) and the position vector $$\backslash mathbf\{r\}$$.

Q3. A current-carrying circular loop of radius R in the x-y plane is partially bent such that its right half (x > 0) now lies in the y-z plane. How does the magnetic moment change?

• A The magnitude of the magnetic moment diminishes.
• B The magnetic moment does not change.
• C The magnitude of B at (0,0,z) for z >> R increases.
• D The magnitude of B at (0,0,z) for z >> R is unchanged.

Explanation: Initially, the magnetic moment is $$M\; =\; I$($\backslash pi$ $R^2$)$ along the z-axis. After bending, the original loop is replaced by two semi-circular loops. The magnetic moment of the semi-circle in the x-y plane is $M' = I($\backslash pi$ $R^2$)/2$ along z-axis, and the moment of the semi-circle in the y-z plane is $M'' = I($\backslash pi$ $R^2$)/2$ along the x-axis. The net magnetic moment is the vector sum $$M\_\{net\}$ = $\backslash sqrt\{M\text{'}^2\; +\; M\text{'}\text{'}^2\}$ = $\backslash sqrt\{2\}$ $\backslash times$ $\backslash frac\{I\; \backslash pi\; R^2\}\{2\}$ = $\backslash frac\{I\; \backslash pi\; R^2\}\{\backslash sqrt\{2\}\}$$. Since $$M\_\{net\}$ < M$, the magnetic moment diminishes.

Q4. An electron moves with uniform velocity along the axis of a current-carrying long solenoid. What happens to the electron?

• A The electron will be accelerated along the axis.
• B The electron path will be circular about the axis.
• C The electron will experience a force at 45° to the axis and execute a helical path.
• D The electron will continue to move with uniform velocity along the axis.

Explanation: A current-carrying solenoid produces a magnetic field primarily along its axis. When an electron moves parallel to this magnetic field (along the axis), the angle $$\backslash theta$$ between its velocity $$\backslash mathbf\{v\}$$ and the magnetic field $$\backslash mathbf\{B\}$$ is $0^$\backslash circ$$. The Lorentz force $$F\; =\; qvB$$\backslash sin$ $\backslash theta$$ is therefore zero ($$F\; =\; evB$$\backslash sin$ 0^$\backslash circ$ = 0$). With no force acting on it, the electron continues to move with its uniform velocity along the axis.

Q5. In a cyclotron, a charged particle:

• A Undergoes acceleration all the time.
• B Speeds up between the dees because of the magnetic field.
• C Speeds up in a dee.
• D Slows down within a dee and speeds up between dees.

Explanation: A charged particle in a cyclotron is accelerated by an electric field applied across the gap between the dees. This electric field causes the particle to gain energy and speed up each time it crosses the gap. The magnetic field, which is perpendicular to the plane of motion, only changes the direction of the velocity, causing the particle to move in a circular path within the dees. Since the particle is accelerated every time it crosses the gap, it undergoes acceleration throughout its journey in the cyclotron.