# A long straight wire AB carries a current of 4 A. A proton P travels at $4\times 10^{6}ms^{-1}$ parallel to the wire 0·2 m from it and in a direction opposite to the current as shown in the figure. Calculate the force which the magnetic field due to the current carrying wire exerts on the proton. Also specify its direction.

The magnetic field at point P due to

Current carrying straight wire AB,

$B=\frac{\mu _{o}}{2\pi }\frac{I}{r}\; \; \; \; \; \; \; (i)$

Where 'I' is current

'r' is the distance at point P from the wire.

Now,

Force acting on moving proton in the magnetic field;

$F=Bqv \; \sin \theta --------(ii)$

Where, 'B' is magnetic field

'q' is charge

'v' is velocity

from equation (i) putting the value of B in equation (ii), we get

$F=\frac{\mu _{o}}{2\pi }\frac{I}{r}q\; v\; \sin \theta$

We have given that;

$v = 4\times 10^{6}m/s$

$q = 1.6\times 10^{-19}C$

$I = 4A$

$r = 0.2m$

Therefore:-

$F=\frac{2\times 10^{-7}\times 4\times 1.6\times 10^{-19}\times 4\times 10^{6}\times \sin 90^{o}}{0.2}\; \; \; \; \; \; \; \;$

$F=\frac{32\times 1.6\times 10^{-20}}{0.2}$

$F=2.56\times 10^{-18}N$

## Related Chapters

### Preparation Products

##### Knockout NEET July 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
##### Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
##### Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Test Series NEET July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-