SATHEE: Magnetics 6 Question 17

Magnetics 6 Question 17

17. A rigid wire loop of square shape having side of length $L$ and resistance $R$ is moving along the $x$ -axis with a constant velocity $v_{0}$ in the plane of the paper. At $t = 0$ , the right edge of the loop enters a region of length $3 L$ , where there is a uniform magnetic field $B_{0}$ into the plane of the paper, as shown in the figure. For sufficiently large $v_{0}$ , the loop eventually crosses the region. Let $x$ be the location of the right edge of the loop. Let $v (x), I (x)$ and $F (x)$ represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of $x$ . Counter-clockwise current is taken as positive.

(2016 Adv.)

Which of the following schematic plot(s) is (are) correct? (Ignore gravity) (a)

(c)

(b)

(d)

Numerical Value

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Solution:

When loop was entering $(x < L)$

$\begin{aligned} φ & = B L x \\ e & = - \frac{d φ}{d t} = - B L \frac{d x}{d t} \\ | e | & = B L v \\ i & = \frac{e}{R} = \frac{B L v}{R} \end{aligned}$

(anti-clockwise)

$F = i l B ($ Left direction $) = \frac{B^{2} L^{2} v}{R} ($ in left direction $)$

$\Rightarrow a = \frac{F}{m} = - \frac{B^{2} L^{2} v}{m R}$

$a = v \frac{d v}{d x}$

$v \frac{d v}{d x} = - \frac{B^{2} L^{2} v}{m R}$

$\Rightarrow \int_{v_{0}}^{v} d v = - \frac{B^{2} L^{2}}{m R} \int_{0}^{x} d x$

$\Rightarrow v = v_{0} - \frac{B^{2} L^{2} v}{m R} x ($ straight line of negative slope for $x < L)$ $I = \frac{B L}{R} v \Rightarrow (I v s x$ will also be straight line of negative slope for $x < L$ )

$\frac{d φ}{d t} = 0, e = 0, i = 0$

$F = 0, x > 4 L$

$e = B l v$

Force also will be in left direction.

$\begin{matrix} i = \frac{B L v}{R} (clockwise) \\ a = \frac{B^{2} L^{2} v}{m R} = v \frac{d v}{d x} \\ F = \frac{B^{2} L^{2} v}{R} \\ \int_{L}^{x} - \frac{B^{2} L^{2}}{m R} d x = \int_{v_{i}}^{v_{f}} d v \\ \Rightarrow - \frac{B^{2} L^{2}}{m R} (x - L) = v_{f} - v_{i} \\ v_{f} - v_{i} - \frac{B^{2} L^{2}}{m R} (x - L) (straight line of negative slope) \\ I = \frac{B L v}{R} \to (Clockwise) (straight line of negative slope) \end{matrix}$

Magnetics 6 Question 17

Numerical Value

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Magnetics 6 Question 17

Numerical Value

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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