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NEET 2021
Exam
NEET
Year
2021
Subject
Physics
Chapter
Magnetic Effects of Current

NEET 2021 · Physics · Magnetic Effects of Current

A thick current carrying cable of radius R carries current I...

A thick current carrying cable of radius RR carries current II uniformly distributed across its cross-section. The variation of magnetic field B(r)B ( r ) due to the cable with the distance rr from the axis of the cable is represented by:A thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. TheA thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. TheA thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. TheA thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. TheA thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. TheA thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. The

Options

  1. AIf 2 ~kg mass is rotating on a circular path of radius 0.8 ~m with angular velocity of 44 rad / sec . If radius of the
  2. BIf 2 ~kg mass is rotating on a circular path of radius 0.8 ~m with angular velocity of 44 rad / sec . If radius of the
  3. CIf 2 ~kg mass is rotating on a circular path of radius 0.8 ~m with angular velocity of 44 rad / sec . If radius of the
  4. DIf 2 ~kg mass is rotating on a circular path of radius 0.8 ~m with angular velocity of 44 rad / sec . If radius of the

Correct answer

C. If 2 ~kg mass is rotating on a circular path of radius 0.8 ~m with angular velocity of 44 rad / sec . If radius of the

Step-by-step solution

A thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. The

For inside part

J=IπR2J = \dfrac{I}{\pi{R}^{2}}

∫B→⋅dl→=μ0Iinclosed…(i)∫ \overset{\to}{B} \cdot \overset{\to}{dl} = {\mu}_{0}{I}_{inclosed} \ldots \left(i\right)

B2\pir=μ0(IπR2)×πr2B2\pir = {\mu}_{0}\left(\dfrac{I}{\pi{R}^{2}}\right) \times \pi{r}^{2}

B2\pir=μ0Ir2R2B2\pir = {\mu}_{0}I\dfrac{{r}^{2}}{{R}^{2}}

B∝rB ∝ r

For outside part

∫B→⋅dl→=μ0Iinclosed∫ \overset{\to}{B} \cdot \overset{\to}{dl} = {\mu}_{0}{I}_{inclosed}

B(2\pir)=μ0IB ( 2\pir ) = {\mu}_{0}I

∴B∝1r∴ B ∝ \dfrac{1}{r}

A thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. The

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