A helium nucleus makes a full rotation in a circle of radius 0.8 meter in 2 sec. The value of the magnetic field induction B in tesla at the centre of circle will be

Question

A helium nucleus makes a full rotation in a circle of radius 0.8 meter in 2 sec. The value of the magnetic field induction B in tesla at the centre of circle will be (A) 2 × 10-19 μ0 (B) 10-19 / μ0 (C) 10-19 μ0 (D) 2 × 10-20 / μ0

Tardigrade Admin · Accepted Answer

B =(μ0/4 π) (2 π i / r ) where i =(2 e / t )=(2 × 1.6 × 10-19/2)=1.6 × 10-19 A ∴ B=(μ0 i/2 r)=(μ0 × 1.6 × 10-19/2 × 0.8)=μ0 × 10-19 T

Q. A helium nucleus makes a full rotation in a circle of radius $0.8$ meter in $2 sec$ . The value of the magnetic field induction $B$ in tesla at the centre of circle will be

$2 \times 1 0^{- 19} μ_{0}$

$1 0^{- 19} / μ_{0}$

$1 0^{- 19} μ_{0}$

$2 \times 1 0^{- 20} / μ_{0}$

$B = \frac{μ _{0}}{4 π} \frac{2 πi}{r}$ where
$i = \frac{2 e}{t} = \frac{2 \times 1.6 \times 1 0 ^{- 19}}{2} = 1.6 \times 1 0^{- 19} A$
$∴ B = \frac{μ _{0} i}{2 r} = \frac{μ _{0} \times 1.6 \times 1 0 ^{- 19}}{2 \times 0.8} = μ_{0} \times 1 0^{- 19} T$

Q. A helium nucleus makes a full rotation in a circle of radius 0.8 meter in 2sec. The value of the magnetic field induction B in tesla at the centre of circle will be

2×10−19μ0​

10−19/μ0​

10−19μ0​

2×10−20/μ0​