Moving Charges and Magnetism - Full Mock Test (Bihar Board/NEET Standard)

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Subjective Section (Answer briefly and clearly):

Q31. State Biot-Savart Law and write its mathematical expression. Explain its significance in magnetism.

Q32. Derive the expression for the magnetic field at the center of a circular coil of radius R carrying current I.

Q33. What is a cyclotron? Write its principle and mention any two limitations.

Q35. Explain Fleming’s Left-Hand Rule with a neat diagram and give one practical application.

Numerical Section:

Q34. A proton moves at 2×10⁶ m/s perpendicular to a magnetic field of 0.2 T. Calculate the force acting on it. (Charge of proton = 1.6×10^-19 C)

Q36. Calculate the magnetic field at the center of a circular loop of radius 0.1 m carrying current 5 A. (μ₀ = 4π × 10^-7 T·m/A)

Q37. A current carrying conductor of length 0.5 m is placed perpendicular to a magnetic field of 0.3 T. Calculate the force on the conductor if current is 4 A.

Q38. Calculate the radius of the circular path of an electron moving at 3×10⁶ m/s in a magnetic field of 0.2 T. (Charge of electron = 1.6×10^-19 C, mass = 9.1×10^-31 kg)

Q39. The magnetic field inside a solenoid is 2×10^-3 T. Calculate the number of turns per meter if the current is 3 A. (μ₀ = 4π × 10^-7 T·m/A)

Subjective Section - Suggested Answers

1. Biot-Savart Law: It gives the magnetic field produced at a point due to a small current element.
   Mathematical Expression:
   dB = (μ₀/4π) × (I·dl × r̂)/r²
   Significance: It is the fundamental law for calculating magnetic field due to any current configuration.
2. Magnetic field at center of a circular coil:
   B = (μ₀I)/(2R)
   Derivation: Using Biot-Savart law, integrate over the loop to get this result.
3. Cyclotron: It is a device to accelerate charged particles using perpendicular electric and magnetic fields.
   Principle: Charged particles move in a circular path in a magnetic field; an alternating electric field accelerates them.
   Limitations: (i) Not suitable for neutral particles, (ii) Relativistic mass increase at high speeds limits acceleration.
4. Fleming’s Left-Hand Rule: If the thumb, forefinger, and middle finger of the left hand are stretched mutually perpendicular, thumb shows force, forefinger magnetic field, and middle finger current.
   Application: Electric motor.
   Diagram: (Draw left hand with fingers as described.)

Numerical Section - Suggested Solutions

1. Q34 Solution:
   F = qvB = 1.6×10^-19 × 2×10⁶ × 0.2 = 6.4×10^-14 N
2. Q36 Solution:
   B = (μ₀I)/(2R) = (4π×10^-7 × 5)/(2 × 0.1) = 3.14×10^-5 T
3. Q37 Solution:
   F = BIL = 0.3 × 4 × 0.5 = 0.6 N
4. Q38 Solution:
   Radius r = (mv)/(qB) = (9.1×10^-31 × 3×10⁶)/(1.6×10^-19 × 0.2) = 8.53×10^-5 m
5. Q39 Solution:
   B = μ₀ n I → n = B/(μ₀ I) = (2×10^-3)/(4π×10^-7 × 3) ≈ 530 turns/m