Question

Each of the two coils has 320 turns. The average radius of the coil is 6 cm. The distance from the center of one coil to the electron beam is 3 cm. If a current of 0.5 amperes runs through the coils, what is the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils

Answer 1

Answer:

Magnetic field, $B=2.39\times 10^{-3}\ T$

Explanation:

It is given that,

Number of turns, N = 320

Radius of the coil, r = 6 cm = 0.06 m

The distance from the center of one coil to the electron beam is 3 cm, x = 3 cm = 0.03 m

Current flowing through the coils, I = 0.5 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :

$B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}$

B = 0.00239 T

or

$B=2.39\times 10^{-3}\ T$

So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is $2.39\times 10^{-3}\ T$. Hence, this is the required solution.