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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Magnetic Field

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Magnetic Field

A metal wire PQ having a mass of 10 g rests on two horizontal metal...

A metal wire PQPQ having a mass of 10 g10\text{ g} rests on two horizontal metal rails separated by 4.90 cm4.90\text{ cm}. A vertically downward magnetic field with a magnitude of 0.800 T0.800\text{ T} is present in the space. As the resistance of the circuit is slowly decreased, it is found that the wire PQPQ begins sliding on the rails once the resistance drops below 20.0 Ω20.0\text{ }\Omega. Determine the coefficient of friction.
A metal wire PQ having a mass of 10 g rests on two horizontal metal rails separated by 4.90 cm . A vertically downward

Options

  1. A0.180.18
  2. B0.240.24
  3. C0.060.06
  4. D0.120.12

Correct answer

D. 0.120.12

Step-by-step solution

When the resistance drops to 20.0 Ω20.0 \text{ }\Omega, the current in the circuit is:

I=VR=620.0=0.3 AI = \dfrac{V}{R} = \dfrac{6}{20.0} = 0.3 \text{ A}

The magnetic force acting on the wire PQPQ is given by:

Fm=ILBF_{m} = I L B

Substituting the given values (L=4.90 cm=4.90×10−2 mL = 4.90 \text{ cm} = 4.90 \times 10^{-2} \text{ m}, B=0.800 TB = 0.800 \text{ T}):

Fm=0.3×(4.90×10−2)×0.800=0.01176 NF_{m} = 0.3 \times (4.90 \times 10^{-2}) \times 0.800 = 0.01176 \text{ N}

The normal force on the wire is equal to its weight (taking g=9.8 m/s2g = 9.8 \text{ m/s}^2):

N=mg=(10×10−3 kg)×9.8 m/s2=0.098 NN = mg = (10 \times 10^{-3} \text{ kg}) \times 9.8 \text{ m/s}^2 = 0.098 \text{ N}

The wire begins to slide when the magnetic force equals the maximum static friction force:

Fm=μNF_{m} = \mu N

μ=FmN=0.011760.098=0.12\mu = \dfrac{F_{m}}{N} = \dfrac{0.01176}{0.098} = 0.12

Therefore, the coefficient of friction is 0.120.12.
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