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Question: An electron experiences a force \((4.0\widehat{i}+3.0\widehat{j})\times {{10}^{-13}}N\) in a uniform...

An electron experiences a force (4.0i^+3.0j^)×1013N(4.0\widehat{i}+3.0\widehat{j})\times {{10}^{-13}}N in a uniform magnetic field when its velocity is 2.5k^×107ms12.5\widehat{k}\times {{10}^{7}}m{{s}^{-1}} . The magnetic field vector B\overrightarrow{B} is:
(A)0.075i^+0.1j^ (B)0.1i^+0.075j^ (C)0.075i^0.1j^+k^ (D)0.075i^0.1j^ \begin{aligned} & (A)-0.075\widehat{i}+0.1\widehat{j} \\\ & (B)0.1\widehat{i}+0.075\widehat{j} \\\ & (C)-0.075\widehat{i}-0.1\widehat{j}+\widehat{k} \\\ & (D)-0.075\widehat{i}-0.1\widehat{j} \\\ \end{aligned}

Explanation

Solution

We know that the force experienced by a charged particle in a magnetic field is given by the product of charge and cross product of velocity and magnetic field. This cross product can be calculated by forming a 3×33\times 3 determinant in X, Y and Z-components of the magnetic field and velocity. And then, equating it with the given force.

Complete answer:
Let the magnetic field vector be given by B\overrightarrow{B}. Then, we can assume this B\overrightarrow{B}as:
B=xi^+yj^+zk^\Rightarrow \overrightarrow{B}=x\widehat{i}+y\widehat{j}+z\widehat{k}
Now, the force on the particle is equal to:
F=e(V×B)\Rightarrow \overrightarrow{F}=e(\overrightarrow{V}\times \overrightarrow{B})
Where,
e is the charge on one electron which is equal to 1.6×1019C-1.6\times {{10}^{-19}}C .
V\overrightarrow{V} is the velocity vector which is given in the problem as:
V=2.5k^×107ms1\Rightarrow \overrightarrow{V}=2.5\widehat{k}\times {{10}^{7}}m{{s}^{-1}}
Putting these values in the above equation we get, the value of force vector as:
F=1.6×1019(2.5k^×107ms1×xi^+yj^+zk^)\Rightarrow \overrightarrow{F}=-1.6\times {{10}^{-19}}(2.5\widehat{k}\times {{10}^{7}}m{{s}^{-1}}\times x\widehat{i}+y\widehat{j}+z\widehat{k})
The cross product in right-hand side of the equation can be solved using the following discriminant as follows:
D=(i^j^k^ 002.5×107 xyz ) D=2.5×107(xj^yi^) \begin{aligned} & \Rightarrow D=\left| \left( \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ 0 & 0 & 2.5\times {{10}^{7}} \\\ x & y & z \\\ \end{matrix} \right) \right| \\\ & \therefore D=2.5\times {{10}^{7}}(x\widehat{j}-y\widehat{i}) \\\ \end{aligned}
Putting this D in the above equation, and using the value of force given in the question, we get:
(4.0i^+3.0j^)×1013N=1.6×1019×2.5×107(xj^yi^)\Rightarrow (4.0\widehat{i}+3.0\widehat{j})\times {{10}^{-13}}N=-1.6\times {{10}^{-19}}\times 2.5\times {{10}^{7}}(x\widehat{j}-y\widehat{i})
On simplifying, we get:
(4.0i^+3.0j^)=(40xj^+40yi^)\Rightarrow (4.0\widehat{i}+3.0\widehat{j})=(-40x\widehat{j}+40y\widehat{i})
On comparing the X and Y components of both the terms in the above equation, we get:
x=340;y=440;z=0 x=0.075;y=0.1;z=0 \begin{aligned} & \Rightarrow x=-\dfrac{3}{40};y=\dfrac{4}{40};z=0 \\\ & \therefore x=-0.075;y=0.1;z=0 \\\ \end{aligned}
Hence, the magnetic field vector influencing the electron comes out to be 0.075i^+0.1j^-0.075\widehat{i}+0.1\widehat{j}.

Hence, option (A) is the correct option.

Note:
These are some basic problems from electromagnetism that require application of basic formulas in steps. But, as we can see, they have somewhat lengthy calculations. So, one should be careful while solving these problems as one mistake at any step can render our entire solution incorrect.