Verified Solution • 2025–26 Session moderate

QPages 249-253: Q34: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q34 Solution: An electron makes revolutions per second in a circle of radius…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the radius from angstroms to meters, using $0.5$ instead of $5 imes 10^{-11}$ m in the denominator of $B = \frac{\mu_0 i}{2r}$, which results in a magnetic field value that is off by a factor of $10^{11}$.

An electron makes 3×1053 \times 10^5 revolutions per second in a circle of radius 0.5 angstrom. Find the magnetic field BB at the centre of the circle.

Step-by-Step Verified Solution

**Answer:** B=12B = 12 T. **Step-by-Step Explanation:** f=3×105f = 3 \times 10^5 rev/s, r=0.5r = 0.5 � =5×10−11= 5 \times 10^{-11} m. i=ef=1.6×10−19×3×105=4.8×10−14i = ef = 1.6 \times 10^{-19} \times 3 \times 10^5 = 4.8 \times 10^{-14} A. B=μ0i2r=4π×10−7×4.8×10−142×5×10−11=6×10−10B = \frac{\mu_0 i}{2r} = \frac{4\pi \times 10^{-7} \times 4.8 \times 10^{-14}}{2 \times 5 \times 10^{-11}} = 6 \times 10^{-10} T.

All Questions in this Chapter

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