Question

A straight wire is placed in a magnetic field that varies with distance $x$ from origin as $B = B_0 (2 - \frac{x}{n}) \hat{k}$. The ends of the wire are at $(a, 0)$ and $(2n, 0)$ and it carries a current $I$. If the force on the wire is $F = 1 B_0 \left(\frac{k a}{2}\right) \hat{j}$, then the value of $k$ is

• A. 1
• B. 5
• C. -1
• D. 1/2

Answer 1

The Correct Option is C

Given: - Magnetic field: \( \vec{B} = B_0 \left(2 - \frac{x}{n} \right) \hat{k} \) - Wire lies along x-axis from \( x = a \) to \( x = 2n \) - Current in wire = \( I \) - Force on a current element in magnetic field: \[ d\vec{F} = I \, d\vec{l} \times \vec{B} \]

Wire description: - \( d\vec{l} = dx \, \hat{i} \) - \( \vec{B} = B_0 \left(2 - \frac{x}{n} \right) \hat{k} \) \[ d\vec{F} = I \, dx \, \hat{i} \times B_0 \left(2 - \frac{x}{n} \right) \hat{k} = I B_0 \left(2 - \frac{x}{n} \right) dx \, (\hat{i} \times \hat{k}) = -I B_0 \left(2 - \frac{x}{n} \right) dx \, \hat{j} \] 

Total force: \[ \vec{F} = \int_a^{2n} d\vec{F} = -I B_0 \int_a^{2n} \left(2 - \frac{x}{n} \right) dx \, \hat{j} \] 

Evaluating the integral: \[ \int_a^{2n} \left(2 - \frac{x}{n} \right) dx = \left[ 2x - \frac{x^2}{2n} \right]_a^{2n} \] \[ = \left(2(2n) - \frac{(2n)^2}{2n} \right) - \left(2a - \frac{a^2}{2n} \right) = (4n - 4n) - \left(2a - \frac{a^2}{2n} \right) = -\left(2a - \frac{a^2}{2n} \right) \] \[ \vec{F} = -I B_0 \left( -2a + \frac{a^2}{2n} \right) \hat{j} = I B_0 \left(2a - \frac{a^2}{2n} \right) \hat{j} \] Given: \[ \vec{F} = I B_0 \left(\frac{k a}{2} \right) \hat{j} \] Equating: \[ 2a - \frac{a^2}{2n} = \frac{k a}{2} \Rightarrow 4a - \frac{a^2}{n} = k a \Rightarrow 4 - \frac{a}{n} = k \] Now, since \( a = 5n \Rightarrow \frac{a}{n} = 5 \), \[ k = 4 - 5 = -1 \] 

Final Answer: –1