A particle having a mass of 0.5 g carries a charge of 2.5× 10-8C . The particle is given an initial horizontal velocity of 6× 104ms-1 . The minimum magnitude of the magnetic field that is required so that, particle will keep moving in a horizontal direction is [take g=10 ms-2 ]

Question

A particle having a mass of 0.5 g carries a charge of 2.5× 10-8C . The particle is given an initial horizontal velocity of 6× 104ms-1 . The minimum magnitude of the magnetic field that is required so that, particle will keep moving in a horizontal direction is [take g=10 ms-2 ] (A)  8.46 T  (B)  3.33 T  (C)  8.64× 10-2T  (D) None of these

Tardigrade Admin · Accepted Answer

To keep the particle moving in horizontal direction

F_{B}

should be equal to weight or

q v B = m g

[For minimum value of

B

angle between

v

and

B

has to be

\frac{π}{2}

]

\Rightarrow

B = \frac{0.5 \times 10 ^{- 3} \times 10}{2.5 \times 1 0 ^{- 8} \times 6 \times 1 0 ^{4}}

= 3.33 T

$8.46 T$

$3.33 T$

$8.64 \times 1 0^{- 2} T$

None of these

To keep the particle moving in horizontal direction $F_{B}$ should be equal to weight or $q v B = m g$
[For minimum value of $B$ angle between $v$ and $B$ has to be $\frac{π}{2}$ ]
$\Rightarrow$ $B = \frac{0.5 \times 10 ^{- 3} \times 10}{2.5 \times 1 0 ^{- 8} \times 6 \times 1 0 ^{4}}$
$= 3.33 T$

Q. A particle having a mass of 0.5g carries a charge of 2.5×10−8C . The particle is given an initial horizontal velocity of 6×104ms−1 . The minimum magnitude of the magnetic field that is required so that, particle will keep moving in a horizontal direction is [take g=10ms−2 ]

8.46T

3.33T

8.64×10−2T

None of these

To keep the particle moving in horizontal direction FB​ should be equal to weight or qvB=mg [For minimum value of B angle between v and B has to be 2π​ ] ⇒ B=2.5×10−8×6×1040.5×10−3×10​ =3.33T

$8.46 T$

$3.33 T$

$8.64 \times 1 0^{- 2} T$

To keep the particle moving in horizontal direction $F_{B}$ should be equal to weight or $q v B = m g$
[For minimum value of $B$ angle between $v$ and $B$ has to be $\frac{π}{2}$ ]
$\Rightarrow$ $B = \frac{0.5 \times 10 ^{- 3} \times 10}{2.5 \times 1 0 ^{- 8} \times 6 \times 1 0 ^{4}}$
$= 3.33 T$