A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2 × 10⁵ ms⁻¹. 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 × 10⁴ N / C. The value of x is _________. Take the mass of the proton =1.6 × 10⁻²⁷ kg.

Solution

Physics sir · 4 steps

Use the undeflected motion to get E = vB, then use the circle to find B.

1. Undeflected motion. With both fields on, the electric and magnetic forces cancel: qE = qvB. So E = vB, and we only need B.
2. Circle gives B. With the electric field off, the magnetic force supplies the centripetal force: qvB = mv²/r. This gives B = mv/qr.

   B = (1.6 × 10⁻²⁷)(2 × 10⁵)/(1.6 × 10⁻¹⁹)(0.02) = (3.2 × 10⁻²²)/(3.2 × 10⁻²¹) = 0.1 T

3. Find E. E = vB = (2 × 10⁵)(0.1) = 2 × 10⁴ N C⁻¹.
4. So. E = 2 × 10⁴ N C⁻¹, so x = 2.

The common slip: 0.02. Using r = 2 instead of 0.02 m gives E = 2 × 10² N C⁻¹, so x = 0.02. Convert cm to m first.

The idea to keep. In crossed fields, the undeflected speed is v = E/B. Switch off E and the radius r = mv/(qB) gives you B. Together they give E = mv²/(qr).

Answer from the NTA final answer key. Solution by PrepWiser.