When there is an electric current through a conducting wire along its length, then an electric field must exist
|
(a) Outside the wire but normal to it |
(b) Outside the wire but parallel to it |
|
(c) Inside the wire but parallel to it |
(d) Inside the wire but normal to it |
(c) Inside the wire but parallel to it
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absorbs an energetic neutron and emits beta particles. The resulting nucleus is
(a) 
|
(b) 
|
(c) 
|
(d) 
|
$\text{m}\omega^2\text{l}$
$\text{Zero.}$
$\text{mg}$
$\frac{1}{2}\text{m}\omega^2\text{l}$
A particle of mass m and charge q enters a magnetic field B perpendicularly with a velocity v, The radius of the circular path described by it will be
|
(a) Bq/mv |
(b) mq/Bv |
(c) mB/qv |
(d) mv/Bq |
Then
