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AP Intermediate 2nd Year Moving Charges and Magnetism — practice questions — page 2

Explore 188 free Moving Charges and Magnetism questions in AP Intermediate 2nd Year Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Moving Charges and Magnetism questions and exercises with answers

Showing questions 31–60 of 188.

A cyclotron of dee radius R accelerates ions of charge q, mass m in a magnetic field B. The maximum kinetic energy isRead the answer and worked solution →A circular loop of radius R carries current I. The magnetic field at a point on its axis at distance x from the centre isRead the answer and worked solution →At the very end of a long solenoid, the magnetic field along the axis isRead the answer and worked solution →A magnetic dipole with moment m makes angle θ with a uniform field B. The torque on the dipole has magnitudeRead the answer and worked solution →An electron moves east in a magnetic field directed north. The magnetic force on the electron pointsRead the answer and worked solution →A cyclotron operates at frequency f in a magnetic field B. The charge-to-mass ratio of the accelerated ion isRead the answer and worked solution →The current sensitivity of a moving coil galvanometer with N turns, coil area A, magnetic field B, spring constant k isRead the answer and worked solution →A 0.1 m radius loop carries 5 A. The magnetic field at a point on its axis 0.1 m from the centre is approximately (μ₀ = 4π × 10⁻⁷ T·m/A)Read the answer and worked solution →A bar magnet with N pole pointing right is placed near a circular current loop whose magnetic moment also points right. The interaction isRead the answer and worked solution →Cyclotrons are not used to accelerate electrons becauseRead the answer and worked solution →A long solenoid has 1000 turns per metre and carries 2 A. The magnetic field strength inside isRead the answer and worked solution →A square coil of side 0.1 m, with 50 turns, carries 1 A in a 0.2 T field. The maximum torque on the coil isRead the answer and worked solution →A charge q moves with velocity v through a region with electric field E and magnetic field B. The net electromagnetic force on it isRead the answer and worked solution →Match each current to its fieldRead the answer and worked solution →Judge these claims on magnetic forceRead the answer and worked solution →The Lorentz forceRead the answer and worked solution →Work done by a magnetic forceRead the answer and worked solution →Reading the Lorentz forceRead the answer and worked solution →Force on a current wireRead the answer and worked solution →The current carrying conductorRead the answer and worked solution →Charge in a uniform fieldRead the answer and worked solution →What the cyclotron frequency ignoresRead the answer and worked solution →The cyclotron machineRead the answer and worked solution →Formula to what it describesRead the answer and worked solution →The Biot-Savart lawRead the answer and worked solution →Using Biot-SavartRead the answer and worked solution →Field of a circular coilRead the answer and worked solution →Odd one out: magnetismRead the answer and worked solution →Depends on speed or notRead the answer and worked solution →True of the magnetic forceRead the answer and worked solution →

Worked practice questions

A cyclotron of dee radius R accelerates ions of charge q, mass m in a magnetic field B. The maximum kinetic energy is

  1. qBR
  2. q²B²R²/(2m)
  3. qBR²/(2m)
  4. q²B²/(2mR)
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Correct answer: B. q²B²R²/(2m)

1. At the edge of the dees, the orbit radius equals R: qvB = mv²/R gives v_max = qBR/m. 2. Kinetic energy K = ½ m v_max². 3. Substitute v_max: K = ½ m (qBR/m)². 4. Simplify: K = q²B²R²/(2m). This is the maximum extracted KE. _Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.4.2_

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A circular loop of radius R carries current I. The magnetic field at a point on its axis at distance x from the centre is

  1. μ₀IR² / 2(R²+x²)^(3/2)
  2. μ₀I(R²+x²)^(1/2) / 2R²
  3. μ₀IR / 2(R²+x²)
  4. μ₀I / 2R
Show answer and solution

Correct answer: A. μ₀IR² / 2(R²+x²)^(3/2)

1. Apply Biot-Savart to each current element on the loop; vertical components cancel by symmetry. 2. Each element contributes dB·sin α along the axis, where sin α = R/(R²+x²)^(1/2). 3. Integrating around the loop gives B = μ₀IR² / (2(R²+x²)^(3/2)). 4. At x = 0 the formula reduces to μ₀I/(2R), as expected for the centre. _Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.6_

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At the very end of a long solenoid, the magnetic field along the axis is

  1. μ₀nI
  2. μ₀nI/2
  3. 2μ₀nI
  4. Zero
Show answer and solution

Correct answer: B. μ₀nI/2

1. The on-axis field of a finite solenoid involves angles subtended at its two ends. 2. Deep inside (both ends far), the formula gives B = μ₀nI. 3. At an end face, one angle is 90° while the other approaches 0°. 4. The field at the end of a long solenoid is therefore B = μ₀nI/2, half the interior value. _Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.8.1_

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