A charge q=-4 μC has an instantaneous velocity 𝐯=(2 𝐢̂-3 𝐣̂+𝐤̂) ×10^6 m / s in a uniform magnet

A charge q=-4 μC has an instantaneous velocity 𝐯=(2 𝐢̂-3...

A charge $q=-4 \mu \mathrm{C}$ has an instantaneous velocity $\mathbf{v}=(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \times 10^{6} \mathrm{~m} / \mathrm{s}$ in a uniform magnetic field $\mathbf{B}=(2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \times 10^{-2} \mathrm{~T}$. What is the force on the charge?

A charge $q=-4 \mu \mathrm{C}$ has an instantaneous velocity $\mathbf{v}=(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \times 10^{6} \mathrm{~m} / \mathrm{s}$ in a uniform
magnetic field $\mathbf{B}=(2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \times 10^{-2} \mathrm{~T}$. What is the force on the charge?

Key Concepts

Lorentz Force Law

This principle states that a moving charge in electromagnetic fields experiences a force given by F = q(v x B), where q is the charge, v is its velocity, and B is the magnetic field. It forms the basis for understanding the interaction between charged particles and magnetic fields.

Vector Cross Product

The cross product of two vectors results in a third vector that is perpendicular to the plane containing the original vectors, with its magnitude determined by the product of their magnitudes and the sine of the angle between them. This operation is crucial for determining both the magnitude and the direction of the magnetic force acting on a moving charge.

Charge Sign and Force Direction

The sign of the charge affects the direction of the force derived from the right-hand rule. For a negative charge, the force direction is opposite to that indicated by the right-hand rule, an important consideration when predicting the motion of charged particles in magnetic fields.

Uniform Magnetic Field

A uniform magnetic field is one that has the same magnitude and direction at every point in space. This consistency simplifies the analysis of the Lorentz force because the magnetic field does not vary spatially, allowing for straightforward calculations using vector operations.

Transcript

00:01 In the given problem, the charge q is having a value of minus 4 microculum.

00:10 It is moving in a magnetic field which is given in vector form as 2 icap plus 5j cap minus 3 k kk x into 10 dash minus 2 tesla.

00:31 And the velocity of this charge particle is given as v is equal to to i cap minus 3 j cap and plus k cap into 10 dash to the part 6 meter per second so here we have to find magnetic lawrence force experienced by this charge particle so this magnet the clorentz force will be given by f is equal to charge q times v cross b in vector of norm so here it will be for charge this is minus 4 into 10 dash minus 6 coulum the cross product of 2 i cap minus 3 j cap plus k cap 2 x xx with 2 i cap plus 5 j cap minus 3 k cap into 10 dash minus 2 so this 10 dash 4 6 this 10 dash 4 it will be 10 dash par plus 4 and here this is or we can say this is 10 dash par minus 6 will be cancelled by 10 to part plus 6 so finally this is minus 4 into 10 dash per minus 2 and here for this cross product we will use matrix method.

02:25 In this matrix method the first row will be having i cap, j cap and k cap.

02:34 Then the second row means the component of factors in this velocity vector component of velocity in this vector.

02:43 So the components are two along x -axis, minus three along y -axis, and just one along z -axis.

02:52 Then the component of magnetic field in the third row, two tesla along x -axis, this is five along y -axis and minus three along z -axis.

03:07 So finally, in order to check the coefficient of this i cap here we will cross multiply minus 3 with minus 3 and 5 with the 1 so it will come out to be f is equal to minus 4 into 10 dash bar minus 2...

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