JEE MAIN - Physics (2004 - No. 27)
The magnetic field due to a current carrying circular loop of radius $$3$$ $$cm$$ at a point on the axis at a distance of $$4$$ $$cm$$ from the centre is $$54\,\mu T.$$ What will be its value at the center of loop?
$$125\,\mu T$$
$$150\,\mu T$$
$$250\,\mu T$$
$$75\,\mu T$$
Explanation
The magnetic field at a point on the axis of a circular loop at a distance $$x$$ from center is,
$$B = {{{\mu _0}i\,{a^2}} \over {2\left( {{x^2} + {a^2}} \right)3/2}}$$ $$\,\,\,\,\,B' = {{{\mu _0}i} \over {2a}}$$
$$\therefore$$ $$B' = {{B.{{\left( {{x^2} + {a^2}} \right)}^{3/2}}} \over {{a^3}}}$$
Put $$x = 4$$ & $$a = 3 \Rightarrow B' = {{54\left( {{5^3}} \right)} \over {3 \times 3 \times 3}} = 250\mu T$$
$$B = {{{\mu _0}i\,{a^2}} \over {2\left( {{x^2} + {a^2}} \right)3/2}}$$ $$\,\,\,\,\,B' = {{{\mu _0}i} \over {2a}}$$
$$\therefore$$ $$B' = {{B.{{\left( {{x^2} + {a^2}} \right)}^{3/2}}} \over {{a^3}}}$$
Put $$x = 4$$ & $$a = 3 \Rightarrow B' = {{54\left( {{5^3}} \right)} \over {3 \times 3 \times 3}} = 250\mu T$$
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