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प्रश्न
A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of 4.0 × 10−4 T parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire.
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उत्तर
Given:
Uniform magnetic field, B0 = 4.0 × 10−4 T
Magnitude of current, I = 30 A
Separation of the point from the wire, d = 0.02 m
Thus, the magnetic field due to current in the wire is given by
\[B = \frac{\mu_0 I}{2\pi d}\]
\[= \frac{2 \times {10}^{- 7} \times 30}{0 . 02}\]
\[ = 3 \times {10}^{- 4} T\]
B0 is perpendicular to B (as shown in the figure).
∴ Resultant magnetic field
\[B_{net} = \sqrt{B^2 + {B_0}^2}\]
\[ = \sqrt{(4 \times {10}^{- 4} )^2 + (3 \times {10}^{- 4} )^2}\]
\[ = 5 \times {10}^{- 4} T\]
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