Solution for Question 14
Step 1: Analyzing Linear Motion (No Magnetic Field)
The equation of motion is $m\frac{dv}{dt} = -bv$.
Rewriting in terms of distance $s$: $mv\frac{dv}{ds} = -bv \implies m dv = -b ds$.
Integrating from initial velocity $v_0$ to stop ($0$):
$$ \int_{v_0}^{0} m \, dv = \int_{0}^{10} -b \, ds $$ $$ -mv_0 = -b(10) \implies mv_0 = 10b \quad \text{…(Equation 1)} $$Step 2: Vector Equation with Magnetic Field
Equation of motion: $m\frac{d\vec{v}}{dt} = q(\vec{v} \times \vec{B}) – b\vec{v}$.
Integrate both sides with respect to time ($dt$):
$$ \int m \, d\vec{v} = \int q(\vec{v} \times \vec{B}) \, dt – \int b\vec{v} \, dt $$Since $\int \vec{v} \, dt = \vec{S}$ (displacement vector):
$$ m(\vec{v}_f – \vec{v}_i) = q(\vec{S} \times \vec{B}) – b\vec{S} $$The particle stops, so $\vec{v}_f = 0$. Let initial velocity be $\vec{v}_i = v_0 \hat{i}$.
$$ -m v_0 \hat{i} = q(\vec{S} \times \vec{B}) – b\vec{S} $$Rearranging:
$$ m v_0 \hat{i} = b\vec{S} – q(\vec{S} \times \vec{B}) $$Step 3: Solving for B
The vector $\vec{S}$ is perpendicular to $\vec{S} \times \vec{B}$. Thus, the magnitudes form a Pythagorean triplet:
$$ (mv_0)^2 = (bS)^2 + |q(\vec{S} \times \vec{B})|^2 $$Since $\vec{v}$ is in the plane and $\vec{B}$ is perpendicular, $|\vec{S} \times \vec{B}| = SB$.
$$ (mv_0)^2 = b^2 S^2 + q^2 S^2 B^2 $$Substitute known values: $mv_0 = 10b$ and $S = 6$.
$$ (10b)^2 = b^2(6)^2 + q^2(6)^2 B^2 $$ $$ 100b^2 = 36b^2 + 36q^2 B^2 $$ $$ 64b^2 = 36q^2 B^2 $$Taking square root:
$$ 8b = 6qB \implies B = \frac{8b}{6q} = \frac{4b}{3q} $$Correct Answer: (a)
$\frac{4b}{3q}$