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Magnetics question

2008 · Shift 0 · Q64
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Magnetics question

2008 · Shift 0 · Q64

JEE MainPhysicsMagneticsMCQ+4 / −1
A horizontal overhead powerline is at height of 4m4m4m from the ground and carries a current of 100A100A100A from east to west. The magnetic field directly below it on the ground is (μ0=4π×10−7  Tm  A−1)\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,\,Tm\,\,{A^{ - 1}}} \right)(μ0​=4π×10−7TmA−1)
  1. A
    2.5×10−7 T2.5 \times {10^{ - 7}}\,T2.5×10−7T southward
  2. B
    5×10−6 T5 \times {10^{ - 6}}\,T5×10−6T northward
  3. C
    5×10−6 T5 \times {10^{ - 6}}\,T5×10−6T southward
  4. D
    2.5×10−7 T2.5 \times {10^{ - 7}}\,T2.5×10−7T northward
View written solutionFree

Correct answer: C

  1. Magnetic field due to a long straight current-carrying wire

For a long straight wire, the magnetic field at perpendicular distance rrr is

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}B=2πrμ0​I​

Here,

  • I=100 AI = 100\,AI=100A
  • r=4 mr = 4\,mr=4m
  • μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\,T\,m\,A^{-1}μ0​=4π×10−7TmA−1
  1. Substitute the values
B=(4π×10−7)(100)2π(4)B = \frac{(4\pi \times 10^{-7})(100)}{2\pi(4)}B=2π(4)(4π×10−7)(100)​

Simplify:

B=4π×10−58π=12×10−5=5×10−6 TB = \frac{4\pi \times 10^{-5}}{8\pi} = \frac{1}{2} \times 10^{-5} = 5 \times 10^{-6}\,TB=8π4π×10−5​=21​×10−5=5×10−6T
  1. Determine the direction

Current flows from east to west.

Using the right-hand thumb rule:

  • Point the thumb toward the west.
  • The curled fingers give the direction of magnetic field around the wire.

At the point directly below the wire (on the ground), the field is toward the south.

  1. Match with the options

Thus the magnetic field is

5×10−6 T5 \times 10^{-6}\,T5×10−6T

in the southward direction.

So the correct option is C.

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