Question

An electron is projected along the positive z-axis, with an initial speed of $5.0\times {10}^{5}m/s$. A uniform magnetic filed is present, but there is no electric field. The electron experiences an initial acceleration whose components are $a_x=-7.0\times {10}^{16}m/s^2,a_y=-3.5\times {10}^{16}m/s^2,a_a=0$. The components of the uniform magnetic field, in $SI$ units, are closest to :

• A. $B_x=0.4$
• B. $B_y=0.6$
• C. $B_z=0.8$
• D. $B_x=0.5$

Answer 1

Answer: (A)

$B_x=0.4$

Given: velocity of electron,$\overrightarrow{v}=5\ast {10}^{5}k$ $m/s$

component of acceleration is,

$a_x=-7\ast {10}^{16}m/s^2$

and $a_y=-3.5\ast {10}^{16}m/s^2$

and $a_z=0m/s^2$

To find: $component$ $of$ $magneticfield$$=?$

Solution: as we know that,

mass of electron,$m=9.1\ast {10}^{-31}kg$

and charge of electron,$e=-1.6\ast {10}^{-19}C$

Now as there is no electric field,

So, $Force,\overrightarrow{F}=e(\overrightarrow{v}\ast \overrightarrow{B})$ $....\left(1\right)$

and also $\overrightarrow{F}=m\ast \overrightarrow{a}$ $....\left(2\right)$

On equating $e{q}^n\left(1\right)ande{q}^n\left(2\right)$,we get,

$m\ast \overrightarrow{a}=e(\overrightarrow{v}\ast \overrightarrow{B})$

Let $\overrightarrow{B}=B_{x}i+B_{y}j+B_{z}k$

$==>9.1\ast {10}^{-31}\ast (-7i-3.5j)\ast {10}^{16}=-1.6\ast {10}^{-19}{.10}^5\left[\right(5k)\ast (B_{x}i+B_{y}j+B_{z}k\left)\right]$

$==>5.6875\ast {10}^{-1}\ast (-7i-3.5j)=-(-5B_{y}i+5B_{x}j)$

$==>0.11375\ast (-7i-3.5j)=(B_{y}i-B_{x}j)$

On comparing both side,we get

$==>B_x=3.5\ast 0.11375$ and $B_y=-7\ast 0.11375$

$==>B_x=0.398125$ and $B_y=-0.79625$

$==>B_x\simeq 0.4$ and$B_y\simeq -0.8$

hence,

The correct opt: A