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प्रश्न
A long, straight wire carrying a current of 1.0 A is placed horizontally in a uniform magnetic field B = 1.0 × 10−5 T pointing vertically upward figure. Find the magnitude of the resultant magnetic field at the points P and Q, both situated at a distance of 2.0 cm from the wire in the same horizontal plane.
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उत्तर
Given:
Uniform magnetic field, B0 = 1.0 × 10−5 T (Vertically upwards)
Separation of the point from the wire, d = 2 cm = 0.02 m
The magnetic field due to the wire is given by
\[B_w = \frac{\mu_0 i}{2\pi d} = \frac{4\pi \times {10}^{- 7} \times 1}{2\pi \times 0 . 02}\]
\[ \Rightarrow B_w = 1 \times {10}^{- 5} T\]
Now,
Net magnetic field at point P:
BP = Bw + B0 = 2 × 10−5 T
Net magnetic field at point Q:
BQ = Bw − B0 = 0
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