Question

solution g A square loop is halfway into a magnetic field that is perpendicular to the plane of the loop. The loop mass is 10 g and its resistance is 0.020 Ω . A switch is closed at t =0s, causing the magnetic field to increase from 0 to 1.0 T in 0.010 s. a) What is the induced current in the square loop? bWith what speed is the loop "kicked" away from the magnetic field? Hint: What is the impulse on the loop?

Answer 1

Answer:

Explanation:

It is given that,

Mass of the loop, m = 10 g = 0.01 kg

Resistance of the loop, R = 0.02 ohms

Dimension of square loop, 7 cm × 7 cm

Area of the loop, $A=49\ cm^2=49\times 10^{-4}\ m^2$

At time t = 0 s, the magnetic field increases from 0 to 1 T in 0.01 s

(a) Due to change in magnetic field, an emf is induced in the loop. Using the formula for induced emf as :

$\epsilon=A\dfrac{dB}{dt}$

$\epsilon=49\times 10^{-4}\times \dfrac{1}{0.01}$

$\epsilon=0.49\ V$

Now using Ohm's law to find the induced current in the loop. It is given by :

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{0.49}{0.02}$

I =24.5 A

(b) A magnetic force acting on the loop is given by :

$F=iLB\ sin\theta$

$F=i(l/2)B$

$F=24.5\times (0.07/2)\times 1$

F = 0.8575

Since, $F=\dfrac{mv}{t}$

$\dfrac{mv}{t}=0.8575$

$v=\dfrac{0.8575\times 0.01}{0.01}$

v = 0.8575 m/s

Hence, this is the require solution.