Class 12 Moving Charges and Magnetism Mock Test (CBSE)

Topics in Moving Charges and Magnetism

• Magnetic force and the Lorentz force
• Motion in a magnetic field
• Magnetic field due to a current element: Biot-Savart law
• Magnetic field on the axis of a circular current loop
• Ampere's circuital law
• The solenoid
• Force between two parallel currents
• Torque on a current loop and magnetic dipole
• The moving coil galvanometer

Moving Charges and Magnetism practice questions with answers

1. Q1Multiple choice·1 mark
   A proton enters a region of uniform magnetic field $\vec{B}$ with velocity $\vec{v}$ perpendicular to $\vec{B}$. The angle between the velocity vector $\vec{v}$ and the magnetic force vector $\vec{F}$ acting on the proton is:

   1. (a)$45^\circ$
   2. (b)$0^\circ$
   3. (c)$180^\circ$
   4. (d)$90^\circ$
   Show answer and step marks

   1. Correct option: (d) $90^\circ$. The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$, which is always perpendicular to both velocity $\vec{v}$ and magnetic field $\vec{B}$, making the angle $90^\circ$.+1

   Answer: (d)
2. Q2Multiple choice·1 mark
   An electron is moving in a circular path of radius $r$ in a uniform magnetic field. If its speed is doubled, the new radius of its circular path will be:

   1. (a)$r$
   2. (b)$r / 2$
   3. (c)$4r$
   4. (d)$2r$
   Show answer and step marks

   1. Correct option: (d) $2r$. The radius of the circular path is $r = \frac{mv}{qB}$, which shows that radius is directly proportional to speed $v$; thus doubling the speed doubles the radius to $2r$.+1

   Answer: (d)
3. Q3Short answer·3 marks
   State Ampere's circuital law. Using this law, derive an expression for the magnetic field at a distance $r$ from a long straight wire carrying a steady current $I$.
   Show answer and step marks

   1. State Ampere's circuital law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$+1
   2. Consider a circular Amperian loop of radius $r$ around the wire where $\vec{B}$ and $d\vec{l}$ are parallel, giving $\oint B \, dl = B(2\pi r)$+1
   3. Equate the line integral to $\mu_0 I$ and solve for $B$: $B = \frac{\mu_0 I}{2\pi r}$+1

   Answer: Ampere's law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$, the line integral of $\vec{B}$ round a closed loop equals $\mu_0$ times the current enclosed; for a long straight wire, $B = \frac{\mu_0 I}{2\pi r}$

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