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A long, straight wire carrying a current of $1.0 A$ is placed horizontally in a uniform magnetic field of $1.0 \times 10^{- 6} T$ pointing vertically upward. 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 [[1]] $\times 10^{- 5}$ $T$ and [[2]]T respectively.

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Detailed Solution

A long, straight wire carrying a current of

1.0 A

is placed horizontally in a uniform magnetic field of

1.0 \times 10^{- 6} T

pointing vertically upward. 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

2 \times 10^{- 5}

T

and

0 T

respectively.
Question Image

Here,

r = 2 \times 1 0^{- 2} m, I = 1 A, B = 10^{- 5} T

At point P,

B_{1} = \frac{μ_{0} I}{2 Πr}

= \frac{2 \times 10^{- 7}}{2 \times 10^{- 2}}

= 10^{- 5} T

.
Therefore net magnetic field at point P

= 10^{- 5} + 1 0^{- 5}

= 2 \times 1 0^{- 5} T

Similarly preceding for point Q, we get

B_{2} = 10^{- 5} T

Since Q is at downward direction therefore net magnetic field at Q is

= 10^{- 5} + (- 1 0^{- 5})

= 0

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