Question

A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes​

Answer 1

Answer:

The radius must be four times

Answer 2

Answer:

r'=4r

Explanation:

Let the velocity of the particle be V

The magnitude of the magnetic field is B

The charge on the particle be q

The mass of the particle is m

The radius of the followed path of the charged particle in the presence of the magnetic field is  

$r=\dfrac{mv}{qB}$

Now the magnitude of the magnetic field increases to $B'=\dfrac{B}{2}$

The velocity of the charged particle is v'=2v

The radius r' followed by the charged particle is

$r'=\dfrac{mv'}{qB'}\\r'=\dfrac{m(2v)}{q\left( \dfrac{B}{2} \right)}\\r'=\dfrac{4mv}{qB}\\r'=4r$

The radius becomes four times