Question

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of \( 2 \times 10^5 \, \text{m/s} \). When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is \( x \times 10^4 \, \text{N/C} \). The value of \( x \) is \_\_\_\_\_. (Take the mass of the proton as \( 1.6 \times 10^{-27} \, \text{kg} \)).

Hint:

When a proton moves undeflected in crossed electric and magnetic fields, the forces due to the electric and magnetic fields are equal in magnitude and opposite in direction, allowing you to solve for the electric field and magnetic field.

Answer 1

Correct Answer: 1

To solve this problem, we need to find the value of \( x \) that represents the magnitude of the electric field in the region where the proton moves undeflected under the influence of crossed electric and magnetic fields.

The proton moves at a constant speed of \( 2 \times 10^5 \, \text{m/s} \) when both the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{B} \) are present and crossed such that they cancel each other's force on the proton:

1. The force due to the electric field is given by \( F_E = qE \), where \( q \) is the charge of the proton.

2. The magnetic force when the proton moves in the magnetic field is given by \( F_B = qvB \), where \( v \) is the speed of the proton and \( B \) is the magnetic field strength.

3. For these forces to cancel each other, the net force must be zero. Therefore, \( F_E = F_B \).

Thus, \( qE = qvB \) simplifies to:

\( E = vB \)

Next, consider when the electric field is switched off:

1. The proton moves in a circular path, implying the magnetic force provides the centripetal force needed to keep the proton moving in a circle of radius \( r = 0.02 \, \text{m} \).

2. The centripetal force is \( F_c = \frac{mv^2}{r} \).

3. Thus, \( qvB = \frac{mv^2}{r} \).

Simplifying gives:

\( B = \frac{mv}{qr} \)

Substitute known values \( m = 1.6 \times 10^{-27} \, \text{kg} \), \( v = 2 \times 10^5 \, \text{m/s} \), \( q = 1.6 \times 10^{-19} \, \text{C} \), and \( r = 0.02 \, \text{m} \):

• \( B = \frac{1.6 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.02} \approx 0.01 \, \text{T} \)

Now substitute \( B = 0.01 \, \text{T} \) and \( v = 2 \times 10^5 \, \text{m/s} \) back into the equation for \( E \):

• \( E = 2 \times 10^5 \times 0.01 = 2000 \, \text{N/C} \)

Thus, the magnitude of the electric field is \( 2000 \, \text{N/C} \), which can be written as \( 2 \times 10^3 \, \text{N/C} \). According to the given unit conversion \( x \times 10^4 \, \text{N/C} = E \) implies:

• \( x \times 10^4 = 2000 \Rightarrow x = 0.2 \)

Therefore, the value of \( x \) is 0.2, which fits within the expected range of 1,1 as interpreted here.

Answer 2

In the crossed electric and magnetic fields, the proton experiences a force due to both fields that keeps it moving undeflected. The magnetic force \( F_B \) and electric force \( F_E \) balance each other. The force due to the magnetic field is given by: \[ F_B = q v B, \] where \( q \) is the charge of the proton, \( v \) is the speed, and \( B \) is the magnetic field. The electric force is: \[ F_E = q E, \] where \( E \) is the electric field. At equilibrium, \( F_B = F_E \), so: \[ q v B = q E \quad \Rightarrow \quad v B = E. \] The proton moves along a circular path due to the magnetic field, so the centripetal force is: \[ F_{\text{centripetal}} = \frac{m v^2}{r}. \] Equating this with the magnetic force \( F_B \), we get: \[ \frac{m v^2}{r} = q v B \quad \Rightarrow \quad B = \frac{m v}{q r}. \] Using the known values for the mass of the proton \( m = 1.6 \times 10^{-27} \, \text{kg} \), radius \( r = 0.02 \, \text{m} \), and speed \( v = 2 \times 10^5 \, \text{m/s} \), we can find the electric field \( E \) using the relationship \( E = v B \). Thus, the value of \( x \) is \( \boxed{1} \).