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QJEEM21Ph1c71: JEE Main 2021 Physics - Class 12 JEEM21Ph1c71 Solution: A coaxial cable consists of an inner wire of radius 'a'…

A coaxial cable consists of an inner wire of radius 'a' surrounded by an outer shell of inner and outer radii 'b' and 'c' respectively. The inner wire carries an electric current i0, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x < a and (ii) a < x < b ?

Multiple Choice Options

  • A x2a2{{{x^2}} \over {{a^2}}}
  • B a2x2{{{a^2}} \over {{x^2}}}
  • C x2b2−a2{{{x^2}} \over {{b^2} - {a^2}}}
  • D b2−a2x2{{{b^2} - {a^2}} \over {{x^2}}}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 27th August Evening Shift Physics - Magnetic Effect of Current Question 107 English Explanation
when x < a

B1(2πx)=μ0(i0πa2)πx2{B_1}(2\pi x) = {\mu _0}\left( {{{{i_0}} \over {\pi {a^2}}}} \right)\pi {x^2}

B(2πx)=μ0i0x2a2B(2\pi x) = {{{\mu _0}{i_0}{x^2}} \over {{a^2}}}

B1=μ0i0x2πa2{B_1} = {{{\mu _0}{i_0}x} \over {2\pi {a^2}}} .... (1)

when a < x < b

B2(2πx)=μ0i0{B_2}(2\pi x) = {\mu _0}{i_0}

B2=μ0i02πx{B_2} = {{{\mu _0}{i_0}} \over {2\pi x}} ..... (2)

B1B2=μ0i0x2πa2μ0i02πx=x2a2{{{B_1}} \over {{B_2}}} = {{{\mu _0}{i_0}{x \over {2\pi {a^2}}}} \over {{{{\mu _0}{i_0}} \over {2\pi x}}}} = {{{x^2}} \over {{a^2}}}

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