A circular coil of radius 40 mm consists of 250 turns of wire in which the current is 20mA. The magnetic field in the center of the coil is μ =4π×10-7Hm-1

Question

A circular coil of radius 40 mm consists of 250 turns of wire in which the current is 20mA. The magnetic field in the center of the coil is μ =4π×10-7Hm-1 (A) 0.785 G (B) 0.525 G (C) 0.629 G (D) 0.900 G

Tardigrade Admin · Accepted Answer

In a circular coil of n turns, magnetic field is

B = \frac{μ _{0} n I}{2 r} = \frac{4 π \times 1 0 ^{- 7} \times 250 \times 20 \times 1 0 ^{- 3}}{2 \times 40 \times 1 0 ^{- 3}}

(∵

n = no. of turns, I = current through
coil, r = radius of coil)

\Rightarrow B = \frac{4 π \times 250 \times 20 \times 1 0 ^{- 7 - 3 + 3}}{2 \times 40}

=250×3.14×

1 0^{- 7}

= 785 \times 1 0^{- 7} = 0.785 \times 1 0^{- 4}

tesla
=0.785 gauss

Q. A circular coil of radius 40 mm consists of 250 turns of wire in which the current is 20mA. The magnetic field in the center of the coil is μ=4π×10−7Hm−1

0.785 G

0.525 G

0.629 G

0.900 G

In a circular coil of n turns, magnetic field is B=2rμ0​nI​=2×40×10−34π×10−7×250×20×10−3​ (∵ n = no. of turns, I = current through coil, r = radius of coil) ⇒B=2×404π×250×20×10−7−3+3​ =250×3.14×10−7 =785×10−7=0.785×10−4tesla =0.785 gauss

Q. A circular coil of radius 40 mm consists of 250 turns of wire in which the current is 20mA. The magnetic field in the center of the coil is $μ = 4 π \times 1 0^{- 7} H m^{- 1}$