Verified Solution • 2025–26 Session moderate

QPages 249-253: Q30: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q30 Solution: A long, straight wire is fixed horizontally and carries a…

A long, straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass density 1.0×10−4 kg m−11.0 \times 10^{-4} \text{ kg m}^{-1} is placed parallel to and directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?

Step-by-Step Verified Solution

**Answer:** i=0.49i = 0.49 A. **Step-by-Step Explanation:** For magnetic repulsion to balance weight: μ0i1i22πd=λg\frac{\mu_0 i_1 i_2}{2\pi d} = \lambda g. 4π×10−7×50×i2π×0.005=1×10−4×9.8\frac{4\pi \times 10^{-7} \times 50 \times i}{2\pi \times 0.005} = 1 \times 10^{-4} \times 9.8. i=10−4×9.8×0.0052×50×10−7=0.49i = \frac{10^{-4} \times 9.8 \times 0.005}{2 \times 50 \times 10^{-7}} = 0.49 A.

All Questions in this Chapter

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