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  • Solve it, - Magnetic Effects of Current and Magnetism - JEE Main-2

 A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B.  The field occupies a region of space by width 'd'. If 'a' be the angle of deviation of proton from initial direction of motion (see figure), the value of \sin \alpha will  be :

  • Option 1)

    \frac{B}{2}\sqrt{\frac{qd}{mV}}

  • Option 2)

    \frac{B}{d}\sqrt{\frac{q}{2mV}}

  • Option 3)

    Bd\sqrt{\frac{q}{2mV}}

  • Option 4)

    qV\sqrt{\frac{Bd}{2m}}

 

Answers (1)

best_answer

As we learnt in

Radius of charged particle -

r=\frac{mv}{qB}=\frac{P}{qB}=\frac{\sqrt{}2mk}{qB}=\frac{1}{B}\sqrt{}\frac{2mV}{q}

-

 

 

Energy of Proton =\frac{1}{2}mv^{2}=qV

v=\sqrt{\frac{2qV}{m}}

Magnetic force qvBsin90^{o}=\frac{mv^{2}}{R}

R=\frac{mv}{qB}

From the figure

sin\alpha=\frac{d}{R}=\frac{qdB}{mv}=\frac{qdB}{m}\sqrt{\frac{m}{2qV}}

\therefore\ sin\alpha=Bd\sqrt{\frac{q}{2mV}}

 


Option 1)

\frac{B}{2}\sqrt{\frac{qd}{mV}}

Incorrect

Option 2)

\frac{B}{d}\sqrt{\frac{q}{2mV}}

Incorrect

Option 3)

Bd\sqrt{\frac{q}{2mV}}

Correct

Option 4)

qV\sqrt{\frac{Bd}{2m}}

Incorrect

Posted by

Aadil

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