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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Magnetic Field due to a Current

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Magnetic Field due to a Current

Determine the magnetic field B at the centre of curvature of a...

Determine the magnetic field BB at the centre of curvature of a semicircular wire that has a radius of 10.0 cm10.0 \text{ cm} and carries a current of 5.0 A5.0 \text{ A}.

Options

  1. A3.14×10−5 T3.14 \times 10^{-5} \text{ T}
  2. B7.85×10−6 T7.85 \times 10^{-6} \text{ T}
  3. C1.57×10−4 T1.57 \times 10^{-4} \text{ T}
  4. D1.57×10−5 T1.57 \times 10^{-5} \text{ T}

Correct answer

D. 1.57×10−5 T1.57 \times 10^{-5} \text{ T}

Step-by-step solution

The magnetic field at the centre of a semicircular wire carrying current II is given by:

B=μ0I4rB = \dfrac{\mu_0 I}{4r}

Given:

I=5.0 AI = 5.0 \text{ A}

r=10.0 cm=0.1 mr = 10.0 \text{ cm} = 0.1 \text{ m}

μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}

Substituting the values into the formula:

B=4π×10−7×5.04×0.1B = \dfrac{4\pi \times 10^{-7} \times 5.0}{4 \times 0.1}

B=π×10−7×5.00.1B = \dfrac{\pi \times 10^{-7} \times 5.0}{0.1}

B=50π×10−7 TB = 50\pi \times 10^{-7} \text{ T}

B=5π×10−6 TB = 5\pi \times 10^{-6} \text{ T}

Substituting π=3.14\pi = 3.14:

B=5×3.14×10−6 TB = 5 \times 3.14 \times 10^{-6} \text{ T}

B=15.7×10−6 TB = 15.7 \times 10^{-6} \text{ T}

B=1.57×10−5 TB = 1.57 \times 10^{-5} \text{ T}
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