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QJEEM13Phbaec: JEE Main 2013 Physics - Class 12 JEEM13Phbaec Solution: Two short bar magnets of length each have magnetic moments and…

Two short bar magnets of length 11 cmcm each have magnetic moments 1.201.20 Am2A{m^2} and 1.001.00 Am2A{m^2} respectively. They are placed on a horizontal table parallel to each other with their NN poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.020.0 cm.cm. The value of the resultant horizontal magnetic induction at the mid-point OO of the line joining their centres is close to ( \left( \, \right. Horizontal component of earth's magnetic induction is 3.6×10.5Wb/m2)3.6 \times 10.5Wb/{m^2})

Multiple Choice Options

  • A 3.6×10.5  Wb/m23.6 \times 10.5\,\,Wb/{m^2}
  • B 2.56×10.4  Wb/m22.56 \times 10.4\,\,Wb/{m^2}
  • C 3.50×10.4  Wb/m23.50 \times 10.4\,\,Wb/{m^2}
  • D 5.80×10.4 Wb/m25.80 \times 10.4\,Wb/{m^2}

Step-by-Step Verified Solution

JEE Main 2013 (Offline) Physics - Magnetic Properties of Matter Question 58 English Explanation

Given: M1=1.20Am2   {M_1} = 1.20A{m^2}\,\,\, and    M2=1.00Am2\,\,\,{M_2} = 1.00A{m^2}

r=202cm=0.1mr = {{20} \over 2}cm = 0.1m

Bnet=B1+B2+BH{B_{net}} = {B_1} + {B_2} + {B_H}

Bnet=μ0(M1+M2)r3+BH{B_{net}} = {{{\mu _0}\left( {{M_1} + {M_2}} \right)} \over {{r^3}}} + {B_H}

=10−7(1.2+1)(0.1)3+3.6×10−5= {{{{10}^{ - 7}}\left( {1.2 + 1} \right)} \over {{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}}

=2.56×10−4  wb/m2= 2.56 \times {10^{ - 4}}\,\,wb/{m^2}

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