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(D) The magnetic field at the center of a circular coil with $N$ turns and radius $a$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2a}$.Sinc

Vedclass · Accepted Answer

$\frac{5}{4} \mu_0$

Vedclass · Answer

(D) The magnetic field at the center of a circular coil with $N$ turns and radius $a$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2a}$.Since the currents are in mutually opposite directions,the net magnetic field at the center is the difference between the fields produced by the two rings: $B = |B_1 - B_2|$.Given: $N = 25$,$I_1 = 0.1 	ext{ A}$,$a_1 = 0.5 	ext{ m}$,$I_2 = 0.2 	ext{ A}$,$a_2 = 2.0 	ext{ m}$.Substituting the values:$B = \frac{\mu_0 N}{2} \left| \frac{I_1}{a_1} - \frac{I_2}{a_2} ight|$$B = \frac{\mu_0 	imes 25}{2} \left| \frac{0.1}{0.5} - \frac{0.2}{2.0} ight|$$B = \frac{25 \mu_0}{2} \left| 0.2 - 0.1 ight|$$B = \frac{25 \mu_0}{2} 	imes 0.1 = \frac{2.5 \mu_0}{2} = \frac{5}{4} \mu_0 	ext{ T}$.

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