A 1 mutext{C} charge is moving with velocity | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.First, calculate the cross product $\vec{v} 	imes \vec{B}

Vedclass · Accepted Answer

$171$

Vedclass · Answer

(B) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.First, calculate the cross product $\vec{v} 	imes \vec{B}$:$\vec{v} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 3 & -5 \end{vmatrix}$$= \hat{i}((-2)(-5) - (3)(3)) - \hat{j}((1)(-5) - (3)(2)) + \hat{k}((1)(3) - (-2)(2))$$= \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)$$= 1\hat{i} + 11\hat{j} + 7\hat{k}$.The magnitude of the cross product is $|\vec{v} 	imes \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$.Given $q = 1 \mu	ext{C} = 10^{-6} 	ext{ C}$, the force magnitude is $F = |q| |\vec{v} 	imes \vec{B}| = 10^{-6} 	imes \sqrt{171} 	ext{ N}$.Comparing this with $\sqrt{\alpha} 	imes 10^{-6} 	ext{ N}$, we get $\alpha = 171$.

Similar Questions

$A$ charge moves in a circular path perpendicular to a magnetic field. The time period of the revolution is independent of

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

$A$ charged particle, when entering a uniform magnetic field, moves in a helical path. If its angular velocity is $4 \pi \times 10^6 \text{ rad s}^{-1}$ and its velocity in the direction of the magnetic field is $3 \times 10^5 \text{ m s}^{-1}$, then the pitch of the helix is: (in $\text{ cm}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module