A particle carrying a charge to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be ( μ 0= permeability for vacuum)

Question

A particle carrying a charge to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be ( μ 0= permeability for vacuum) (A) (10-7/μ0) (B) 10-17 μ0 (C) 10-6 μ0 (D) 10-7 μ0

Tardigrade Admin · Accepted Answer

Current, i=(q/t)=100× e Bcentre=(μ0/4π). (2π i/r) =(μ0/4π). (2π×100e/e) (μ0×200×1.6×10-19/4×0.8) =10-17 μ0

Q. A particle carrying a charge to $100$ times the charge on an electron is rotating per second in a circular path of radius $0.8 m$ . The value of the magnetic field produced at the centre will be
( $μ_{0} =$ permeability for vacuum)

$\frac{1 0 ^{- 7}}{μ _{0}}$

$1 0^{- 17} μ_{0}$

$1 0^{- 6} μ_{0}$

$1 0^{- 7} μ_{0}$

Current, $i = \frac{q}{t} = 100 \times e$
$B_{ce n t re} = \frac{μ _{0}}{4 π} . \frac{2 πi}{r}$
$= \frac{μ _{0}}{4 π} . \frac{2 π \times 100 e}{e}$
$\frac{μ _{0} \times 200 \times 1.6 \times 1 0 ^{- 19}}{4 \times 0.8}$
$= 1 0^{- 17} μ_{0}$

Q. A particle carrying a charge to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8m. The value of the magnetic field produced at the centre will be ( μ0​= permeability for vacuum)

μ0​10−7​

10−17μ0​

10−6μ0​

10−7μ0​