Physics-12 : 4: Moving Charges and Magnetism - Flashcards
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Does the magnetic force do work on a charged particle?
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No. The magnetic force is always perpendicular to the velocity.
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| # | Question | Answer |
|---|---|---|
| 1 | Does the magnetic force do work on a charged particle? | No. The magnetic force is always perpendicular to the velocity. |
| 2 | What is the magnitude of the magnetic force on a charged particle? | \( F = q v B \sin\theta \). |
| 3 | What is the angular frequency of a charged particle's circular motion in a magnetic field? | \( \omega = \frac{qB}{m} \). |
| 4 | What is 1 gauss in tesla? | \( 10^{-4} \) T. |
| 5 | What is the magnetic field at the centre of a circular loop carrying current \( I \)? | \( B = \frac{\mu_0 I}{2R} \). |
| 6 | An element \( \Delta\mathbf{l} = \Delta x \hat{i} \) at the origin carries a current of 10 A. What is the magnetic field on the y-axis at 0.5 m? (\( \Delta x = 1 \) cm) | \( 4 \times 10^{-8} \) T in the +z direction. |
| 7 | A tightly wound 100 turn coil of radius 10 cm carries a current of 1 A. What is the magnetic field at the centre? | \( 6.28 \times 10^{-4} \) T. |
| 8 | Do the straight segments of a semicircular wire contribute to the magnetic field at the centre? | No. |
| 9 | What is the magnetic field inside a long solenoid? | \( B = \mu_0 n I \). |
| 10 | How does the magnetic field vary with distance inside a current-carrying wire? | Directly proportional to the distance from the axis. |
| 11 | What is the definition of the ampere? | The steady current which, when maintained in each of two long parallel conductors placed 1 m apart in vacuum, produces a force of \( 2 \times 10^{-7} \) N per metre of length. |
| 12 | What is the force on a wire carrying current from south to north in the earth's magnetic field? | Zero. |
| 13 | When is the torque on a current loop zero? | When the plane of the loop is perpendicular to the magnetic field (\( \theta = 0^\circ \)). |
| 14 | A coil with magnetic moment 10 A m² is placed in a magnetic field of 2 T. What is the torque when the coil has rotated by 90°? | 20 N m. |
| 15 | How is a galvanometer converted into a voltmeter? | By connecting a high resistance in series with it. |
| 16 | Is a current loop equivalent to a magnetic dipole? | Yes, at large distances. |
| 17 | What is the torque on a current loop? | \( \boldsymbol{\tau} = \mathbf{m} \times \mathbf{B} \). |
| 18 | A closely wound solenoid 80 cm long has 5 layers of 400 turns each and carries 8 A. What is the magnetic field inside? | \( 2.5 \times 10^{-2} \) T. |
| 19 | What is the magnetic field 1.5 m below a horizontal power line carrying 90 A from east to west? | \( 1.2 \times 10^{-5} \) T (southward). |
| 20 | What kind of source produces the electrostatic field, and what kind produces the magnetic field? | The electrostatic field is produced by a scalar source (electric charge); the magnetic field is produced by a vector source (the current element \( Id\mathbf{l} \)). |
| 21 | A flexible, irregularly-shaped current loop sits in an external magnetic field. Why does it tend to reshape itself into a circle? | To maximise the flux through it — for a given perimeter, a circle encloses the greatest possible area of any shape, and the loop naturally orients and deforms to maximise flux. |
| 22 | If the number of turns N in a galvanometer coil is doubled, what happens to its current sensitivity and its voltage sensitivity? | Current sensitivity doubles (it's proportional to N). Voltage sensitivity stays unchanged, because the coil's resistance R also roughly doubles (being proportional to wire length), and voltage sensitivity depends on N/R. |
| 23 | What is the approximate magnitude of the Earth's magnetic field? | About \( 3.6 \times 10^{-5} \) T. |
| 24 | For the same two moving coil meters, what is the ratio of voltage sensitivity of \( M_2 \) to \( M_1 \)? | 1 (equal voltage sensitivity). |
| 25 | A galvanometer of resistance 100 Ω gives full-scale deflection for a current of 1 mA. What shunt resistance converts it into an ammeter reading up to 1 A? | ≈ 0.1 Ω — from \( r_s = \frac{I_g G}{I - I_g} = \frac{0.001 \times 100}{1 - 0.001} \), using the fact that the galvanometer and shunt carry the same voltage (Ohm's law on the parallel combination), the same principle used in Example 4.12. |