Question

किसी प्रोटॉन को उसके वेग की दिशा के अभिलम्बवत् एकसमान चुम्बकीय क्षेत्र में प्रवेश कराने
से पूर्व विभवान्तर V तक त्वरित किया गया । यदि विभवान्तर को दुगुना कर दिया जाए, तो
चुम्बकीय क्षेत्र में प्रोटॉन द्वारा चले गए वृत्ताकार पथ की त्रिज्या किस प्रकार परिवर्तित होगी?
A proton is accelerated through a potential difference V, subjected to a
uniform magnetic field acting normal to the velocity of the proton. If the
potential difference is doubled, how will the radius of the circular path
described by the proton in the magnetic field change?​

Answer 1

$\mathfrak{ \huge{ \red{ \underline{ \underline{Answer}}}}} \\ \\ \star \: \bold{ \blue{given}}$

=> A proton is accelerated through potential difference V, subjected to uniform magnetic field acting normal to the velocity of proton.

$\star \: \bold{ \blue{to \: find}}$

=> what is radius of circular path if potential difference will be doubled..

$\star \: \bold{ \blue{formula}}$

=> radius of circular path in forms of mass(m),potential difference(V),charge(q) and magnetic field(B) is given by...

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{ \pink{r = \frac{ \sqrt{2 \times m \times V} }{q \times b}}}}$

=> Here mass of proton, charge of proton, and magnetic field are constant...

=> So, radius of circular path depends only upon potential diifference...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{ \pink{r \propto \: \sqrt{V}}}}$

$\star \: \bold{ \blue{calculation}}$

$\implies \: \frac{r1}{r2} = \frac{ \sqrt{V1} }{ \sqrt{V2} } \\ \\ \implies \: here \: V2 = 2V1 \\ \\ \implies \: \frac{r1}{r2} = \frac{1}{ \sqrt{2} } \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge{\star } \: \: \: \boxed{ \boxed{ \huge \green{r2 = \sqrt{2} r1 }}} \: \: \: \huge{ \star}$