A helium nucleus makes full rotation in a circle of radius 0.8 min 2 s. The value of'magnetic field B at the centre of the circle, will be (μ 0= permeability constant)

Question

A helium nucleus makes full rotation in a circle of radius 0.8 min 2 s. The value of'magnetic field B at the centre of the circle, will be (μ 0= permeability constant) (A)  (2 × 10-19/μ 0)  (B) 2 × 10-19 μ 0  (C) 10-19 μ 0  (D)  (10-19/μ0)

Tardigrade Admin · Accepted Answer

The magnetic field at the centre of a circle is given by

B = \frac{μ _{0} i}{2 r}

where, i is current and r the radius of circle.
Also,

i = \frac{q}{t}

For helium nucleus, q = 2e

∴ i = \frac{2 e}{t}

So,

B = \frac{μ _{0} .2 e}{2 r t}

= \frac{μ _{0} \times 2 \times 1.6 \times 1 0 ^{- 19}}{2 \times 0.8 \times 2}

= 1 0^{- 19} μ_{0}

Q. A helium nucleus makes full rotation in a circle of
radius 0.8 min 2 s. The value of'magnetic field B at the
centre of the circle, will be $(μ_{0} =$ permeability constant)

$\frac{2 \times 1 0 ^{- 19}}{μ _{0}}$

$2 \times 1 0^{- 19} μ_{0}$

$1 0^{- 19} μ_{0}$

$\frac{1 0 ^{- 19}}{μ _{0}}$

None of these

Q. A helium nucleus makes full rotation in a circle of radius 0.8 min 2 s. The value of'magnetic field B at the centre of the circle, will be (μ0​= permeability constant)

μ0​2×10−19​

2×10−19μ0​

10−19μ0​

μ0​10−19​