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QJEEM22Phe88b: JEE Main 2022 Physics - Class 12 JEEM22Phe88b Solution: The magnetic field at the center of current carrying circular…

The magnetic field at the center of current carrying circular loop is B1B_{1} . The magnetic field at a distance of 3\sqrt{3} times radius of the given circular loop from the center on its axis is B2B_{2} . The value of B1/B2B_{1} / B_{2} will be

Multiple Choice Options

  • A 9:04
  • B 12 : 5\sqrt5
  • C 8:01
  • D 5 : 3\sqrt3

Step-by-Step Verified Solution

B1=μ0i2R{B_1} = {{{\mu _0}i} \over {2R}}

B2=μ0iR22(R2+x2)32{B_2} = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {x^2})}^{{3 \over 2}}}}}

⇒B1B2=1R3(R2+x2)32\Rightarrow {{{B_1}} \over {{B_2}}} = {1 \over {{R^3}}}{({R^2} + {x^2})^{{3 \over 2}}}

=1R3(8R3)= {1 \over {{R^3}}}(8{R^3})

=8= 8

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