Imaging that an electron revolves around a circle of the radius 5.3× 10- 11 m with a linear speed of 7.5 × 104 m s- 1 in a hydrogen atom. The magnetic field produced at the centre of the circle, due to the electron, is

Question

Imaging that an electron revolves around a circle of the radius 5.3× 10- 11 m with a linear speed of 7.5 × 104 m s- 1 in a hydrogen atom. The magnetic field produced at the centre of the circle, due to the electron, is (A) 43 Wbm- 2 (B) 4300 Wbm- 2 (C) 0.43 Wb m- 2 (D) 43× 10- 4 Wbm- 2

Tardigrade Admin · Accepted Answer

B=(μ 0/4 π )(e v/r2) =(10-7 × 1.6 × 10-19 × 7.5 × 10+4/(5.3 × 10-11)2) On solving, B=0.43 Wb m- 2

Q. Imaging that an electron revolves around a circle of the radius $5.3 \times 1 0^{- 11} m$ with a linear speed of $7.5 \times 1 0^{4} m s^{- 1}$ in a hydrogen atom. The magnetic field produced at the centre of the circle, due to the electron, is

$43 Wb m^{- 2}$

$4300 Wb m^{- 2}$

$0.43 Wb m^{- 2}$

$43 \times 1 0^{- 4} Wb m^{- 2}$

$B = \frac{μ _{0}}{4 π} \frac{e v}{r ^{2}}$
$= \frac{1 0 ^{- 7} \times 1.6 \times 1 0 ^{- 19} \times 7.5 \times 1 0 ^{+ 4}}{( 5.3 \times 1 0 ^{- 11} ) ^{2}}$
On solving, $B = 0.43 Wb m^{- 2}$

Q. Imaging that an electron revolves around a circle of the radius 5.3×10−11m with a linear speed of 7.5×104ms−1 in a hydrogen atom. The magnetic field produced at the centre of the circle, due to the electron, is

43Wbm−2

4300Wbm−2

0.43Wbm−2

43×10−4Wbm−2