Question

A charge particle moving in circular path of radius
R produces magnetic field B at the centre of circle.
If it is moving with uniform speed v, then v is
proportional to
1. BR
2. B²R
3. R²/B
4. R²B​

Answer 1

Explanation:

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 : ... Moving Charges and Magnetism ... So, Bqv=rmv2 or r=Bqmv

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Answer 2

Answer: The correct answer is option $1$.

Explanation:

The force $(F_{B} )$ acting on a charged particle $(q)$ moving with velocity $(v)$ in a circular path of radius $(R)$ produces a uniform magnetic field $(B)$ is given by:

$F_{B} = qvB$

This force is balanced by equal and opposite centripetal force acting on the charged particle of mass $(m)$  in a circular path of radius $(R)$ as:

$F_{c}= \frac{mv^{2} }{R}$

$F_{B} = F_{C}$

$qvB = \frac{mv^{2} }{R}$

It gives the following relation:

$v = \frac{BqR}{m}$

Now as the charge and mass of the charged particle is constant therefore velocity is proportional to:

$v$ ∝ $BR$

Therefore correct option is $1$.

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