A straight wire carrying a current of 12 A is sent into a semicircular loop of radius 2.0 cm. What will be magnetic field at the centre of the semicircular loop?

Question

A straight wire carrying a current of 12 A is sent into a semicircular loop of radius 2.0 cm. What will be magnetic field at the centre of the semicircular loop? (A) B =1.9 × 10-4 T (B) B =1.9 × 104 T (C) B =1.9 × 10-4 T, otimes (D) B =1.9 × 104 T, otimes

Tardigrade Admin · Accepted Answer

B=(μ0/4 π) (π i/R)=10-7 × (22 × 12/7 × 2 × 10-2) =1.9 × 10-4 T Thus, B is 1.9 × 10-4 T normal to the plane of the paper going into it.

Q. A straight wire carrying a current of $12 A$ is sent into a semicircular loop of radius $2.0 c m$ . What will be magnetic field at the centre of the semicircular loop?

$B = 1.9 \times 1 0^{- 4} T$

$B = 1.9 \times 1 0^{4} T$

$B = 1.9 \times 1 0^{- 4} T, \otimes$

$B = 1.9 \times 1 0^{4} T, \otimes$

$B = \frac{μ _{0}}{4 π} \frac{πi}{R} = 1 0^{- 7} \times \frac{22 \times 12}{7 \times 2 \times 1 0 ^{- 2}}$
$= 1.9 \times 1 0^{- 4} T$
Thus, $B$ is $1.9 \times 1 0^{- 4}$ T normal to the plane of the paper going into it.

Q. A straight wire carrying a current of 12A is sent into a semicircular loop of radius 2.0cm. What will be magnetic field at the centre of the semicircular loop?

B=1.9×10−4T

B=1.9×104T

B=1.9×10−4T,⊗

B=1.9×104T,⊗