Question

1. A particle of charge +7.5 μC and a speed of 32.5 m/s enters a uniform magnetic field whose magnitude is 0.50 T. Find the magnitude and direction of the magnetic force on the particle.

2. A uniform magnetic field of magnitude 2 G is directed in the +x direction on the x-y plane. A proton enters the magnetic field in the +y direction with a speed of 2 x 10⁶ m/s. What is the magnitude and direction of the magnetic force of the proton? (1T = 10,000 G)

3. A circular coil 50 cm in diameter is rotating in a magnetic field directed upward with a magnitude of 65 milliTesla. Calculate the magnetic flux through the coil at a) horizontal position and b) tilted 60^o from the horizontal position.

Answer 1

1. Given,

$q=+7.5\mu C$

$v=32.5 m/s$

$B=0.50 T$

$F=qv\times B$

Now, substituting the values,

$\Rightarrow F= 7.5\times 32.5\times 0.50N$

$\Rightarrow F = 121.875N$

2.Given,

Magnitude of uniform magnetic field $(B)=2G=2\times 10^{-4}T$

Speed of the proton $(v)=2 \times10^6 m/s$

Charge on the proton $(q)=1.6\times 10^{-19}C$

Magnetic force on the proton $(F)=qv\times B$

$=1.6\times 10^{-19}\times2\times 10^{6}\times2\times 10^{-4}N$

$=6.4\times 10^{-17}N$

As per the left hand thumb rule, direction of force will be along to the z axis.

3.Given,

Diameter of the circular coil (d)=50cm = 0.5m

Magnitude of magnetic field $(B)=65mT= 6.5\times 10^{-2}T$

a) Magnetic flux $(\phi) = B.A = B. \pi( \dfrac{d}{2})^2$

$=6.5\times 10^{-2}\times 3.14\times \frac{1}{16}$

$=1.27\times 10^{-2}Wb$

When tilted at $60^\circ$

$\Rightarrow \phi = BA \cos 60^\circ$

$\Rightarrow \phi = 1.27\times 10^{-2}\times\frac{1}{2}$

$\Rightarrow \phi = 0.635\times 10^{-2}Wb$