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Question: An electron moving with a velocity $\vec{v_1}=2\hat{i}$ $m/s$ at a point in a magnetic field experie...

An electron moving with a velocity v1⃗=2i^\vec{v_1}=2\hat{i} m/sm/s at a point in a magnetic field experiences a force F1⃗=−2j^N\vec{F_1}=-2\hat{j}N. If the electron is moving with a velocity v2⃗=2j^\vec{v_2}=2\hat{j} m/sm/s at the same point, it experiences a force F2⃗=+2i^N\vec{F_2}=+2\hat{i}N. The force the electron would experience if it were moving with a velocity v3⃗=2k^\vec{v_3}=2\hat{k} m/sm/s at the same point, is:

A

0 N

Answer

0N

Explanation

Solution

The force experienced by a charged particle moving in a magnetic field is given by the Lorentz force formula: F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B})

For an electron, the charge q=−eq = -e. So, the formula becomes: F⃗=−e(v⃗×B⃗)\vec{F} = -e(\vec{v} \times \vec{B})

Let the magnetic field at the point be B⃗=Bxi^+Byj^+Bzk^\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}.

Scenario 1: Given: v1⃗=2i^\vec{v_1}=2\hat{i} m/s and F1⃗=−2j^\vec{F_1}=-2\hat{j} N. Substituting these into the Lorentz force formula: −2j^=−e((2i^)×(Bxi^+Byj^+Bzk^))-2\hat{j} = -e((2\hat{i}) \times (B_x\hat{i} + B_y\hat{j} + B_z\hat{k})) −2j^=−2e(i^×Bxi^+i^×Byj^+i^×Bzk^)-2\hat{j} = -2e(\hat{i} \times B_x\hat{i} + \hat{i} \times B_y\hat{j} + \hat{i} \times B_z\hat{k}) Using the cross product rules (i^×i^=0\hat{i} \times \hat{i} = 0, i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j}): −2j^=−2e(0+Byk^−Bzj^)-2\hat{j} = -2e(0 + B_y\hat{k} - B_z\hat{j}) −2j^=−2eByk^+2eBzj^-2\hat{j} = -2eB_y\hat{k} + 2eB_z\hat{j}

Comparing the coefficients of j^\hat{j} and k^\hat{k}: For j^\hat{j}: −2=2eBz  ⟹  eBz=−1-2 = 2eB_z \implies eB_z = -1 (Equation 1) For k^\hat{k}: 0=−2eBy  ⟹  eBy=00 = -2eB_y \implies eB_y = 0 (Equation 2)

Scenario 2: Given: v2⃗=2j^\vec{v_2}=2\hat{j} m/s and F2⃗=+2i^\vec{F_2}=+2\hat{i} N. Substituting these into the Lorentz force formula: 2i^=−e((2j^)×(Bxi^+Byj^+Bzk^))2\hat{i} = -e((2\hat{j}) \times (B_x\hat{i} + B_y\hat{j} + B_z\hat{k})) 2i^=−2e(j^×Bxi^+j^×Byj^+j^×Bzk^)2\hat{i} = -2e(\hat{j} \times B_x\hat{i} + \hat{j} \times B_y\hat{j} + \hat{j} \times B_z\hat{k}) Using the cross product rules (j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}, j^×j^=0\hat{j} \times \hat{j} = 0, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}): 2i^=−2e(−Bxk^+0+Bzi^)2\hat{i} = -2e(-B_x\hat{k} + 0 + B_z\hat{i}) 2i^=2eBxk^−2eBzi^2\hat{i} = 2eB_x\hat{k} - 2eB_z\hat{i}

Comparing the coefficients of i^\hat{i} and k^\hat{k}: For i^\hat{i}: 2=−2eBz  ⟹  eBz=−12 = -2eB_z \implies eB_z = -1 (Equation 3) - This is consistent with Equation 1. For k^\hat{k}: 0=2eBx  ⟹  eBx=00 = 2eB_x \implies eB_x = 0 (Equation 4)

Determine the magnetic field B⃗\vec{B}: From Equations 1, 2, and 4, we have: eBx=0  ⟹  Bx=0eB_x = 0 \implies B_x = 0 eBy=0  ⟹  By=0eB_y = 0 \implies B_y = 0 eBz=−1  ⟹  Bz=−1/eeB_z = -1 \implies B_z = -1/e

So, the magnetic field at that point is B⃗=−1ek^\vec{B} = -\frac{1}{e}\hat{k}.

Scenario 3: We need to find the force F3⃗\vec{F_3} if the electron is moving with a velocity v3⃗=2k^\vec{v_3}=2\hat{k} m/s. F3⃗=−e(v3⃗×B⃗)\vec{F_3} = -e(\vec{v_3} \times \vec{B}) F3⃗=−e((2k^)×(−1ek^))\vec{F_3} = -e((2\hat{k}) \times (-\frac{1}{e}\hat{k})) F3⃗=−e×2×(−1e)(k^×k^)\vec{F_3} = -e \times 2 \times (-\frac{1}{e}) (\hat{k} \times \hat{k}) F3⃗=2(k^×k^)\vec{F_3} = 2 (\hat{k} \times \hat{k}) Since the cross product of any vector with itself is zero (k^×k^=0\hat{k} \times \hat{k} = 0): F3⃗=2×0=0\vec{F_3} = 2 \times 0 = 0 N

The force experienced by the electron is 00 N. This is expected because the electron's velocity is anti-parallel to the magnetic field, and the magnetic force is zero when the velocity is parallel or anti-parallel to the magnetic field.