A conducting wire carrying a steady current $I$ is shaped as shown in the figure below. All connecte...

The magnetic moment of a current-carrying loop can be found by breaking the loop into sub-loops because of the superposition principle, where the total magnetic moment is the vector sum of the moments of the sub-loops. This approach simplifies calculations, especially for loops with complex shapes, by allowing the use of simpler geometric sub-loops. The total magnetic moment is given by $\mathrm{m}_{\text {total }}=I \cdot \mathbf{A}_{\text {total }}$, where the area vector of the entire loop is the sum of the area vectors of the sub-loops, since the same current flows through all parts.$$ \begin{gathered} \mathbf{m}_{\text {total }}=I \cdot\left(\vec{A}_{\text {total }}\right)=I \cdot\left(\vec{A}_1+\vec{A}_2\right) \\ \mathbf{m}_{\text {total }}=I \cdot(a b(\hat{k})+a b(\hat{\jmath})) \\ \mathbf{m}_{\text {total }}=I a b(\hat{k}+\hat{\jmath}) \end{gathered} $$

A particle of mass $m$ and charge $q$ moving with a velocity $\vec{v}=v_0(\hat{i}+\hat{j}-\hat{k})$ ...

The motion of a charged particle in a uniform magnetic field can be decomposed into two independent motions:Motion parallel to the magnetic field, which is uniform. This component determines the pitch ($p$) of the helical trajectory.Motion perpendicular to the magnetic field, which is circular. This component determines the radius ($r$) of the helical trajectory.Let the given velocity be $\vec{v}=v_0(\hat{i}+\hat{j}-\hat{k})$ and the magnetic field be $\vec{B}=B_0(\hat{i}+\hat{j}+\hat{k})$.First, we need to find the components of the velocity vector parallel ($v_{\parallel}$) and perpendicular ($v_{\perp}$) to the magnetic field.1. Calculate the magnitude of the magnetic field and velocity:$|\vec{B}| = \sqrt{(B_0)^2 + (B_0)^2 + (B_0)^2} = \sqrt{3B_0^2} = B_0\sqrt{3}$$|\vec{v}| = \sqrt{(v_0)^2 + (v_0)^2 + (-v_0)^2} = \sqrt{3v_0^2} = v_0\sqrt{3}$2. Calculate the component of velocity parallel to the magnetic field ($v_{\parallel}$):The dot product of $\vec{v}$ and $\vec{B}$ is:$\vec{v} \cdot \vec{B} = [v_0(\hat{i}+\hat{j}-\hat{k})] \cdot [B_0(\hat{i}+\hat{j}+\hat{k})]$$\vec{v} \cdot \vec{B} = v_0 B_0 (1 \cdot 1 + 1 \cdot 1 + (-1) \cdot 1) = v_0 B_0 (1+1-1) = v_0 B_0$The magnitude of the velocity component parallel to $\vec{B}$ is $v_{\parallel} = \frac{|\vec{v} \cdot \vec{B}|}{|\vec{B}|}$:$v_{\parallel} = \frac{v_0 B_0}{B_0\sqrt{3}} = \frac{v_0}{\sqrt{3}} = \frac{\sqrt{3}v_0}{3}$3. Calculate the component of velocity perpendicular to the magnetic field ($v_{\perp}$):We know that $|\vec{v}|^2 = v_{\parallel}^2 + v_{\perp}^2$.So, $v_{\perp}^2 = |\vec{v}|^2 - v_{\parallel}^2$$v_{\perp}^2 = (v_0\sqrt{3})^2 - \left(\frac{v_0}{\sqrt{3}}\right)^2$$v_{\perp}^2 = 3v_0^2 - \frac{v_0^2}{3} = \frac{9v_0^2 - v_0^2}{3} = \frac{8v_0^2}{3}$$v_{\perp} = \sqrt{\frac{8v_0^2}{3}} = \frac{2\sqrt{2}v_0}{\sqrt{3}} = \frac{2\sqrt{2}v_0\sqrt{3}}{3} = \frac{2\sqrt{6}v_0}{3}$4. Calculate the radius of the helical trajectory ($r$):The magnetic force provides the centripetal force for the circular motion.$q v_{\perp} |\vec{B}| = \frac{m v_{\perp}^2}{r}$$r = \frac{m v_{\perp}}{q |\vec{B}|}$Substitute the values of $v_{\perp}$ and $|\vec{B}|$:$r = \frac{m \left(\frac{2\sqrt{6}v_0}{3}\right)}{q (B_0\sqrt{3})}$$r = \frac{2\sqrt{6} m v_0}{3 q B_0 \sqrt{3}}$$r = \frac{2\sqrt{2}\sqrt{3} m v_0}{3 q B_0 \sqrt{3}}$$r = \frac{2\sqrt{2} m v_0}{3 q B_0}$5. Calculate the pitch of the helical trajectory ($p$):The pitch is the distance traveled along the magnetic field direction in one period of circular motion.First, calculate the period $T$ of the circular motion:$T = \frac{2\pi m}{q |\vec{B}|}$$T = \frac{2\pi m}{q B_0\sqrt{3}}$Now, calculate the pitch $p = v_{\parallel} T$:$p = \left(\frac{\sqrt{3}v_0}{3}\right) \left(\frac{2\pi m}{q B_0\sqrt{3}}\right)$$p = \frac{\sqrt{3} v_0 (2\pi m)}{3 q B_0 \sqrt{3}}$$p = \frac{2\pi m v_0}{3 q B_0}$Comparing our calculated values with the given options:$r = \frac{2\sqrt{2} m v_0}{3 q B_0}$$p = \frac{2\pi m v_0}{3 q B_0}$These values match Option A.The final answer is $\boxed{\text{Option A}}$

Moving Charges And Magnetism - IAT-IISER Physics PYQs

The motion of a charged particle in a uniform magnetic field can be decomposed into two independent motions:Motion parallel to the magnetic field, which is uniform. This component determines the pitch (

p

) of the helical trajectory.Motion perpendicular to the magnetic field, which is circular. This component determines the radius (

r

) of the helical trajectory.Let the given velocity be

%5Cvec%7Bv%7D%3Dv_0(%5Chat%7Bi%7D%2B%5Chat%7Bj%7D-%5Chat%7Bk%7D)

and the magnetic field be

%5Cvec%7BB%7D%3DB_0(%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%2B%5Chat%7Bk%7D)

.First, we need to find the components of the velocity vector parallel (

v_%7B%5Cparallel%7D

) and perpendicular (

v_%7B%5Cperp%7D

) to the magnetic field.1. Calculate the magnitude of the magnetic field and velocity:

%7C%5Cvec%7BB%7D%7C%20%3D%20%5Csqrt%7B(B_0)%5E2%20%2B%20(B_0)%5E2%20%2B%20(B_0)%5E2%7D%20%3D%20%5Csqrt%7B3B_0%5E2%7D%20%3D%20B_0%5Csqrt%7B3%7D

%7C%5Cvec%7Bv%7D%7C%20%3D%20%5Csqrt%7B(v_0)%5E2%20%2B%20(v_0)%5E2%20%2B%20(-v_0)%5E2%7D%20%3D%20%5Csqrt%7B3v_0%5E2%7D%20%3D%20v_0%5Csqrt%7B3%7D

2. Calculate the component of velocity parallel to the magnetic field (

v_%7B%5Cparallel%7D

):The dot product of

%5Cvec%7Bv%7D

and

%5Cvec%7BB%7D

is:

%5Cvec%7Bv%7D%20%5Ccdot%20%5Cvec%7BB%7D%20%3D%20%5Bv_0(%5Chat%7Bi%7D%2B%5Chat%7Bj%7D-%5Chat%7Bk%7D)%5D%20%5Ccdot%20%5BB_0(%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%2B%5Chat%7Bk%7D)%5D

%5Cvec%7Bv%7D%20%5Ccdot%20%5Cvec%7BB%7D%20%3D%20v_0%20B_0%20(1%20%5Ccdot%201%20%2B%201%20%5Ccdot%201%20%2B%20(-1)%20%5Ccdot%201)%20%3D%20v_0%20B_0%20(1%2B1-1)%20%3D%20v_0%20B_0

The magnitude of the velocity component parallel to

%5Cvec%7BB%7D

v_%7B%5Cparallel%7D%20%3D%20%5Cfrac%7B%7C%5Cvec%7Bv%7D%20%5Ccdot%20%5Cvec%7BB%7D%7C%7D%7B%7C%5Cvec%7BB%7D%7C%7D

v_%7B%5Cparallel%7D%20%3D%20%5Cfrac%7Bv_0%20B_0%7D%7BB_0%5Csqrt%7B3%7D%7D%20%3D%20%5Cfrac%7Bv_0%7D%7B%5Csqrt%7B3%7D%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B3%7Dv_0%7D%7B3%7D

3. Calculate the component of velocity perpendicular to the magnetic field (

v_%7B%5Cperp%7D

):We know that

%7C%5Cvec%7Bv%7D%7C%5E2%20%3D%20v_%7B%5Cparallel%7D%5E2%20%2B%20v_%7B%5Cperp%7D%5E2

.So,

v_%7B%5Cperp%7D%5E2%20%3D%20%7C%5Cvec%7Bv%7D%7C%5E2%20-%20v_%7B%5Cparallel%7D%5E2

v_%7B%5Cperp%7D%5E2%20%3D%20(v_0%5Csqrt%7B3%7D)%5E2%20-%20%5Cleft(%5Cfrac%7Bv_0%7D%7B%5Csqrt%7B3%7D%7D%5Cright)%5E2

v_%7B%5Cperp%7D%5E2%20%3D%203v_0%5E2%20-%20%5Cfrac%7Bv_0%5E2%7D%7B3%7D%20%3D%20%5Cfrac%7B9v_0%5E2%20-%20v_0%5E2%7D%7B3%7D%20%3D%20%5Cfrac%7B8v_0%5E2%7D%7B3%7D

v_%7B%5Cperp%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B8v_0%5E2%7D%7B3%7D%7D%20%3D%20%5Cfrac%7B2%5Csqrt%7B2%7Dv_0%7D%7B%5Csqrt%7B3%7D%7D%20%3D%20%5Cfrac%7B2%5Csqrt%7B2%7Dv_0%5Csqrt%7B3%7D%7D%7B3%7D%20%3D%20%5Cfrac%7B2%5Csqrt%7B6%7Dv_0%7D%7B3%7D

4. Calculate the radius of the helical trajectory (

r

):The magnetic force provides the centripetal force for the circular motion.

q%20v_%7B%5Cperp%7D%20%7C%5Cvec%7BB%7D%7C%20%3D%20%5Cfrac%7Bm%20v_%7B%5Cperp%7D%5E2%7D%7Br%7D

r%20%3D%20%5Cfrac%7Bm%20v_%7B%5Cperp%7D%7D%7Bq%20%7C%5Cvec%7BB%7D%7C%7D

Substitute the values of

v_%7B%5Cperp%7D

and

%7C%5Cvec%7BB%7D%7C

r%20%3D%20%5Cfrac%7Bm%20%5Cleft(%5Cfrac%7B2%5Csqrt%7B6%7Dv_0%7D%7B3%7D%5Cright)%7D%7Bq%20(B_0%5Csqrt%7B3%7D)%7D

r%20%3D%20%5Cfrac%7B2%5Csqrt%7B6%7D%20m%20v_0%7D%7B3%20q%20B_0%20%5Csqrt%7B3%7D%7D

r%20%3D%20%5Cfrac%7B2%5Csqrt%7B2%7D%5Csqrt%7B3%7D%20m%20v_0%7D%7B3%20q%20B_0%20%5Csqrt%7B3%7D%7D

r%20%3D%20%5Cfrac%7B2%5Csqrt%7B2%7D%20m%20v_0%7D%7B3%20q%20B_0%7D

5. Calculate the pitch of the helical trajectory (

p

):The pitch is the distance traveled along the magnetic field direction in one period of circular motion.First, calculate the period

T

of the circular motion:

T%20%3D%20%5Cfrac%7B2%5Cpi%20m%7D%7Bq%20%7C%5Cvec%7BB%7D%7C%7D

T%20%3D%20%5Cfrac%7B2%5Cpi%20m%7D%7Bq%20B_0%5Csqrt%7B3%7D%7D

Now, calculate the pitch

p%20%3D%20v_%7B%5Cparallel%7D%20T

p%20%3D%20%5Cleft(%5Cfrac%7B%5Csqrt%7B3%7Dv_0%7D%7B3%7D%5Cright)%20%5Cleft(%5Cfrac%7B2%5Cpi%20m%7D%7Bq%20B_0%5Csqrt%7B3%7D%7D%5Cright)

p%20%3D%20%5Cfrac%7B%5Csqrt%7B3%7D%20v_0%20(2%5Cpi%20m)%7D%7B3%20q%20B_0%20%5Csqrt%7B3%7D%7D

p%20%3D%20%5Cfrac%7B2%5Cpi%20m%20v_0%7D%7B3%20q%20B_0%7D

Comparing our calculated values with the given options:

r%20%3D%20%5Cfrac%7B2%5Csqrt%7B2%7D%20m%20v_0%7D%7B3%20q%20B_0%7D

p%20%3D%20%5Cfrac%7B2%5Cpi%20m%20v_0%7D%7B3%20q%20B_0%7D

These values match Option A.The final answer is

%5Cboxed%7B%5Ctext%7BOption%20A%7D%7D

Moving Charges And Magnetism

High-Yield Trend

Questions
2 MCQs

Official Solution

Official Solution

Moving Charges And Magnetism

High-Yield Trend

Questions 2 MCQs

Official Solution

Official Solution

Questions
2 MCQs