Question

a Helium nucleus makes a full rotation in a circle of radius 0.8 m in 2s the value of the magnetic field will be​

Answer 1

Correct option is

B

10

−19

μ

0

The current due to the revolution of the He nucleus is I=

2π

qω

=

2

2×1.6×10

−19

=1.6×10

−19

A

Magnetic field at the center B=

2r

μ

0

I

=

2×0.8

μ

0

×1.6×10

−19

=10

−19

μ

0

Answer 2

Answer:

mark brainliest bro

Explanation

magnetic field that is B=μ0I2r. Now for solving this, we need to find the value of I, for that, we use the equation I=eω2π

then using the values of r,ω, and e that are given in the question we can easily find the answer to the question.

Complete Step-by-Step solution:

Here we are asked to find the magnetic field B at the center of the circle as shown in figure 1.

Figure 1

So we know that the equation of the magnetic field is given by

B=μ0I2r------------------------------------------- (1)

Here μ0 is the permeability of free space

I is the current due to the revolution of the electrons

r=0.8m is the radius of the circle given in the question

For solving equation (1) we need to know the value of I

We know that a current I is always flowing in the nth orbit of an atom due to its electrons revolution, and the equation corresponding to that is given by

I=eω2π

--------------------------------------------- (2)

Here, e=2×1.6×10−19C is the charge of a He atom.

ω is the angular frequency of the nth orbit.

Given that time is equal to 2 second, so from this we can get the frequency as f=1t=12Hz

Now we can also write f=ω2π=12------------------------- (3)

Substituting equation (3) and value of e in (2) we get

ForI=eω2π=2×1.6×10−192=1.6×10−19A

----------------------- (4)

Now substituting equation (4) in equation (1) we get

B=μ01.6×10−192×0.8=10−19μ0

hope I helped you