A particle of mass m and charge q has an initial velocity \overrightarrow v = v_0\widehat j . If an electric field \overrightarrow E = E_0\widehat i and magnetic field \overrightarrow B = B_0\widehat i act on the particle, its speed will double after a time:

Magnetics PYQ — JEE Main 2020 · 7 Jan · Shift 2 · Q56 with Solution

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Correct answer: C

Given data

Initial velocity: $\vec v(0)=v_0\hat j$
Electric field: $\vec E=E_0\hat i$
Magnetic field: $\vec B=B_0\hat i$

We need the time when the speed becomes double, i.e. $|ec v|=2v_0.$

Write Lorentz force equation

The force on the particle is $m\frac{d\vec v}{dt}=q(\vec E+\vec v\times \vec B).$

Let $\vec v=v_x\hat i+v_y\hat j+v_z\hat k.$ Since $\vec B=B_0\hat i$ ,

\begin{vmatrix} \hat i & \hat j & \hat k\\ v_x & v_y & v_z\\ B_0 & 0 & 0 \end{vmatrix} = v_z B_0\hat j - v_y B_0\hat k.$$ So, $$m\frac{d\vec v}{dt}=q\left(E_0\hat i+v_zB_0\hat j-v_yB_0\hat k\right).$$ Thus the component equations are: $$m\frac{dv_x}{dt}=qE_0,$$ $$m\frac{dv_y}{dt}=qB_0v_z,$$ $$m\frac{dv_z}{dt}=-qB_0v_y.$$ --- 3. **Observe the role of magnetic field** The magnetic force is always perpendicular to velocity, so it does **no work**. Hence only the electric field changes the kinetic energy. Using power: $$\frac{d}{dt}\left(\frac12 mv^2\right)=q\vec E\cdot \vec v=qE_0v_x.$$ But from $$m\frac{dv_x}{dt}=qE_0,$$ with initial condition $v_x(0)=0$, we get $$v_x=\frac{qE_0}{m}t.$$ So the electric field continuously increases the $x$-component of velocity, while the magnetic field only rotates the $y$-$z$ part. --- 4. **Magnitude of the $y$-$z$ part remains constant** From the equations, $$\frac{d}{dt}(v_y^2+v_z^2)=2v_y\frac{dv_y}{dt}+2v_z\frac{dv_z}{dt}$$ $$=2v_y\left(\frac{qB_0}{m}v_z\right)+2v_z\left(-\frac{qB_0}{m}v_y\right)=0.$$ Hence, $$v_y^2+v_z^2=\text{constant}.$$ Initially, $v_y(0)=v_0$ and $v_z(0)=0$, so $$v_y^2+v_z^2=v_0^2.$$ Therefore total speed is $$v^2=v_x^2+v_y^2+v_z^2=v_x^2+v_0^2.$$ Since $$v_x=\frac{qE_0}{m}t,$$ we have $$v^2=v_0^2+\left(\frac{qE_0}{m}t\right)^2.$$ --- 5. **Condition for speed to double** We want $$v=2v_0.$$ So, $$(2v_0)^2=v_0^2+\left(\frac{qE_0}{m}t\right)^2,$$ $$4v_0^2=v_0^2+\left(\frac{qE_0}{m}t\right)^2,$$ $$\left(\frac{qE_0}{m}t\right)^2=3v_0^2.$$ Thus, $$t=\frac{\sqrt{3}mv_0}{qE_0}.$$ --- 6. **Match with options** This corresponds to: **Option C:** $$\boxed{\frac{\sqrt{3}mv_0}{qE_0}}$$ --- 7. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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View written solutionFree

Correct answer: C

Given data

Initial velocity: $\vec v(0)=v_0\hat j$
Electric field: $\vec E=E_0\hat i$
Magnetic field: $\vec B=B_0\hat i$

We need the time when the speed becomes double, i.e. $|ec v|=2v_0.$

Write Lorentz force equation

The force on the particle is $m\frac{d\vec v}{dt}=q(\vec E+\vec v\times \vec B).$

Let $\vec v=v_x\hat i+v_y\hat j+v_z\hat k.$ Since $\vec B=B_0\hat i$ ,

\begin{vmatrix} \hat i & \hat j & \hat k\\ v_x & v_y & v_z\\ B_0 & 0 & 0 \end{vmatrix} = v_z B_0\hat j - v_y B_0\hat k.$$ So, $$m\frac{d\vec v}{dt}=q\left(E_0\hat i+v_zB_0\hat j-v_yB_0\hat k\right).$$ Thus the component equations are: $$m\frac{dv_x}{dt}=qE_0,$$ $$m\frac{dv_y}{dt}=qB_0v_z,$$ $$m\frac{dv_z}{dt}=-qB_0v_y.$$ --- 3. **Observe the role of magnetic field** The magnetic force is always perpendicular to velocity, so it does **no work**. Hence only the electric field changes the kinetic energy. Using power: $$\frac{d}{dt}\left(\frac12 mv^2\right)=q\vec E\cdot \vec v=qE_0v_x.$$ But from $$m\frac{dv_x}{dt}=qE_0,$$ with initial condition $v_x(0)=0$, we get $$v_x=\frac{qE_0}{m}t.$$ So the electric field continuously increases the $x$-component of velocity, while the magnetic field only rotates the $y$-$z$ part. --- 4. **Magnitude of the $y$-$z$ part remains constant** From the equations, $$\frac{d}{dt}(v_y^2+v_z^2)=2v_y\frac{dv_y}{dt}+2v_z\frac{dv_z}{dt}$$ $$=2v_y\left(\frac{qB_0}{m}v_z\right)+2v_z\left(-\frac{qB_0}{m}v_y\right)=0.$$ Hence, $$v_y^2+v_z^2=\text{constant}.$$ Initially, $v_y(0)=v_0$ and $v_z(0)=0$, so $$v_y^2+v_z^2=v_0^2.$$ Therefore total speed is $$v^2=v_x^2+v_y^2+v_z^2=v_x^2+v_0^2.$$ Since $$v_x=\frac{qE_0}{m}t,$$ we have $$v^2=v_0^2+\left(\frac{qE_0}{m}t\right)^2.$$ --- 5. **Condition for speed to double** We want $$v=2v_0.$$ So, $$(2v_0)^2=v_0^2+\left(\frac{qE_0}{m}t\right)^2,$$ $$4v_0^2=v_0^2+\left(\frac{qE_0}{m}t\right)^2,$$ $$\left(\frac{qE_0}{m}t\right)^2=3v_0^2.$$ Thus, $$t=\frac{\sqrt{3}mv_0}{qE_0}.$$ --- 6. **Match with options** This corresponds to: **Option C:** $$\boxed{\frac{\sqrt{3}mv_0}{qE_0}}$$ --- 7. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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