# A charged particle is accelerated through a potential difference of $24\,kV$ and acquires a speed of $2\times {10^6}\,m/s$. It is then injected perpendicularly into a magnetic field of strength $0.2\,T$. Find the radius of the circle described by it.

Answer

Verified

278.7k+ views

**Hint:**To solve this question, we need to look for using the concept of centripetal force and apply it along with the concept of force on a charge. The concept of conservation of energy is also useful in this problem.

**Complete step by step answer:**

It is given that the charged particle is accelerated through a potential difference of $24\times{10^3}\,V$.And as a result it acquires a speed of $2\times {10^6}\,m/s$. Now from the question we can easily understand that the centripetal force will be equal to the force on the charge and we can use this equation for further parts of the question. Therefore,

$\dfrac{m{{v}^{2}}}{r}=qVB$ ,

simplifying this equation we get the following

$r=\dfrac{mv}{qB}$..........................$\left( 1 \right)$

Moving ahead, with the help of conservation of energy theorem, we can say that the kinetic energy of the charge initially will be equal to the potential energy of the charge after it injects into the magnetic field. Therefore, according to conservation of energy theorem,

$\dfrac{1}{2}m{{v}^{2}}=q{{V}_{B}}$

After simplifying, we get that

$\dfrac{1}{q}=\dfrac{2{{V}_{B}}}{m{{v}^{2}}}$.................$\left( 2 \right)$

Now using both of the equation and putting the value of $\dfrac{1}{q}$ in the first equation

$r=\dfrac{2{{V}_{B}}}{m{{v}^{2}}}\times \dfrac{mv}{B}$

we can get that

$r=\dfrac{2{{V}_{B}}}{BV}$

Now, putting the values we get the answer

\[r=\dfrac{\left( 2\times 24 \times{10^3}\right)}{\left( 2\times 0.2\times {{10}^{6}} \right)}\\

\Rightarrow r=\dfrac{48\times{10^3}}{0.4\times{10^6}}\\

\Rightarrow r=0.120\,m\\

\therefore r=12\,cm\]

**Therefore the radius of the circle described in the magnetic field is $12\,cm$.**

**Note:**The question says that the charged particle is inserted perpendicularly, so we solve the question as above, but if the question had any other conditions like the charge is inserted parallel or at a given angle then the solution would have changed slightly.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Define absolute refractive index of a medium

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Trending doubts

The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE