BITSAT 2015 Paper

As we know F = qvB =

mv2

r

∴ r =

mv

Bq

And KE = k =

1

2

mv2
∴ mv = √2km
∴ r =

mv

qB

=

√2km

qB

⇒ r α √k or r = c1∕2 (c is a constant)

dr

r

= c

dk1∕2

dr

or

cΔk

Δr

= 2 √k
or

Δr

r

=

cΔk

2√kc√k

=

Δk

2k

Therefore percentage changes in radius ofpath,

Δr

r

× 100 =

Δk

2k

× 100 = 2%