
In a certain region, static electric and magnetic fields exist. The magnetic field is given by $\vec{B} = B_0 \left(\hat{i} + 2\hat{j}-4\hat{k}\right)$. If a test charge given by $\vec{v} = v_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right)$ experiences no force in that region, then the electric field in that region, in SI units is:
A. $\vec{E} = -v_0B_0\left(\hat{i} +\hat{j}-7\hat{k}\right)$
B. $\vec{E} = -v_0B_0\left(14\hat{j}+7\hat{k}\right)$
C. $\vec{E} = v_0B_0\left(14\hat{j}+7\hat{k}\right)$
D. $\vec{E} = -v_0B_0\left(3\hat{i} -2\hat{j}-4\hat{k}\right)$
Answer
449.1k+ views
Hint: Both the electric field and the magnetic field exert a force on the test charge and influence its trajectory. In such a case, arrive at the resultant expression for the net force acting on the test charge, which we know to be the Lorentz force. Recall that if a test charge experiences no force, it traverses through the electromagnetic field at a constant velocity. To this end, calculate the appropriate magnitude of the existing electric field .
Formula used:
Lorentz force: $\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right)$
Complete step-by-step answer:
We know that electric and magnetic fields are invisible fields of force that are responsible for influencing charged particles that traverse through them and define their trajectory. The electric and magnetic forces thus exerted on the particle have a resultant effect on the particle’s motion through the electromagnetic space.
This resultant force acting on a charged particle moving under the influence of the electric and magnetic forces is known as the Lorentz force. If a particle of charge$\;q$ moving with a velocity $\vec{v}$ in and electric field $\vec{E}$ and magnetic field $\vec{B}$, the Lorentz force $\vec{F}$ experienced by it is given as:
$\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right)$
In the context of the question, we are given that:
$\vec{B} = B_0 \left(\hat{i} + 2\hat{j}-4\hat{k}\right)$ and $\vec{v} = v_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right)$
And the charged particle experiences no force in the electromagnetic region, i.e.,
$\vec{F} = 0$
$\Rightarrow q\left(\vec{E} + (\vec{v} \times \vec{B})\right) = 0$
$\Rightarrow \vec{E} + (\vec{v} \times \vec{B}) = 0$
$\Rightarrow \vec{E} = -(\vec{v} \times \vec{B})$
Plugging in the given values we get:
$\vec{E} = - \left(v_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right)\right) \times \left(B_0 \left(\hat{i} + 2\hat{j}-4\hat{k}\right)\right)$
$\Rightarrow \vec{E} = -v_0B_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right) \times \left(\hat{i} + 2\hat{j}-4\hat{k}\right)$
$\Rightarrow \vec{E} = -v_0B_0\left(-\hat{k} -2\hat{j} -6\hat{k}+4\hat{i}-12\hat{j}-4\hat{i}\right) = -v_0B_0\left(+4\hat{i}-4\hat{i}-2\hat{j}-12\hat{j}-\hat{k}-6\hat{k}\right)$
$\Rightarrow \vec{E} = -v_0B_0 \left(0\hat{i}-14\hat{j}-7\hat{k}\right) = v_0B_0\left(14\hat{j} + 7\hat{k}\right)$
So, the correct answer is “Option C”.
Note: A common misconception is that in the Lorentz force expression, the velocity or trajectory of the charged particle and the magnetic field that it traverses through must be perpendicular to each other for a finite magnetic force to act on it. This, however, is not the case, since the cross product between$\vec{v}$ and $\vec{B}$ suggests that:
$\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right) = q\left(\vec{E} + (|\vec{v}|| \vec{B}|\;sin\theta)\right)$, where $\theta$ can be any angle between $\vec{v}$ and $\vec{B}$.
Additionally, the Lorentz force acting on the particle will always be perpendicular to the plane containing $\vec{v}$ and $\vec{B}$.
Formula used:
Lorentz force: $\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right)$
Complete step-by-step answer:
We know that electric and magnetic fields are invisible fields of force that are responsible for influencing charged particles that traverse through them and define their trajectory. The electric and magnetic forces thus exerted on the particle have a resultant effect on the particle’s motion through the electromagnetic space.
This resultant force acting on a charged particle moving under the influence of the electric and magnetic forces is known as the Lorentz force. If a particle of charge$\;q$ moving with a velocity $\vec{v}$ in and electric field $\vec{E}$ and magnetic field $\vec{B}$, the Lorentz force $\vec{F}$ experienced by it is given as:
$\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right)$
In the context of the question, we are given that:
$\vec{B} = B_0 \left(\hat{i} + 2\hat{j}-4\hat{k}\right)$ and $\vec{v} = v_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right)$
And the charged particle experiences no force in the electromagnetic region, i.e.,
$\vec{F} = 0$
$\Rightarrow q\left(\vec{E} + (\vec{v} \times \vec{B})\right) = 0$
$\Rightarrow \vec{E} + (\vec{v} \times \vec{B}) = 0$
$\Rightarrow \vec{E} = -(\vec{v} \times \vec{B})$
Plugging in the given values we get:
$\vec{E} = - \left(v_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right)\right) \times \left(B_0 \left(\hat{i} + 2\hat{j}-4\hat{k}\right)\right)$
$\Rightarrow \vec{E} = -v_0B_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right) \times \left(\hat{i} + 2\hat{j}-4\hat{k}\right)$
$\Rightarrow \vec{E} = -v_0B_0\left(-\hat{k} -2\hat{j} -6\hat{k}+4\hat{i}-12\hat{j}-4\hat{i}\right) = -v_0B_0\left(+4\hat{i}-4\hat{i}-2\hat{j}-12\hat{j}-\hat{k}-6\hat{k}\right)$
$\Rightarrow \vec{E} = -v_0B_0 \left(0\hat{i}-14\hat{j}-7\hat{k}\right) = v_0B_0\left(14\hat{j} + 7\hat{k}\right)$
So, the correct answer is “Option C”.
Note: A common misconception is that in the Lorentz force expression, the velocity or trajectory of the charged particle and the magnetic field that it traverses through must be perpendicular to each other for a finite magnetic force to act on it. This, however, is not the case, since the cross product between$\vec{v}$ and $\vec{B}$ suggests that:
$\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right) = q\left(\vec{E} + (|\vec{v}|| \vec{B}|\;sin\theta)\right)$, where $\theta$ can be any angle between $\vec{v}$ and $\vec{B}$.
Additionally, the Lorentz force acting on the particle will always be perpendicular to the plane containing $\vec{v}$ and $\vec{B}$.
Recently Updated Pages
Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Write the following in Roman numerals 25819 class 7 maths CBSE

Trending doubts
Give 10 examples of unisexual and bisexual flowers

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

What are the major means of transport Explain each class 12 social science CBSE

Franz thinks Will they make them sing in German even class 12 english CBSE
