An electron enters a uniform magnetic field B = 0.23 T at an angle of 45° to the magnetic field (see accompanying figure). Determine the radius r and the pitch p (distance between loops) of the electron's helical path, assuming its speed is 3.0 x 106 m/s.

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An electron enters a uniform magnetic field B = 0.23 T at an angle of 45° to the magnetic field (see accompanying figure). Determine the radius r and the pitch p (distance between loops) of the electron&#039;s helical path, assuming its speed is 3.0 x 106 m/s.

user_27qwe · Accepted Answer

Given information:Magnetic field, B=0.23TAngle between the velocity and magnetic field, θ=45∘Speed of the electron, v=3.0×106m/sFirst, let&#039;s find the radius, r, of the electron&#039;s helical path. In circular motion, the Lorentz force is responsible for the centripetal force acting on the charged particle.The Lorentz force, F, can be expressed as:F=q(v×B)where q is the charge of the electron, v is the velocity of the electron, and B is the magnetic field.The centripetal force, Fc, is given by:Fc=mv2rwhere m is the mass of the electron and r is the radius of the helical path.Setting F=Fc, we have:q(v×B)=mv2rSince the angle between v and B is 45∘, we can write v×B as:v×B=vBsin(45∘)Substituting this into the equation, we get:q(vBsin(45∘))=mv2rSimplifying, we find:r=mvqBsin(45∘)Substituting the given values into the equation:m=9.1×10−31kgv=3.0×106m/sq=−1.6×10−19CB=0.23Twe can calculate the radius r:r=(9.1×10−31kg)(3.0×106m/s)(−1.6×10−19C)(0.23T)(sin(45∘))Calculating this expression, we find:r≈3.14×10−3mTherefore, the radius of the electron&#039;s helical path is approximately 3.14×10−3 meters.Next, let&#039;s find the pitch, p, of the electron&#039;s helical path. The pitch is the distance between successive loops along the helical path.The pitch p can be calculated using the formula:p=2πrsin(θ)Substituting the given values into the equation:r=3.14×10−3mθ=45∘we can calculate the pitch p:p=2π(3.14×10−3m)sin(45∘)Calculating this expression, we find:p≈0.0089mTherefore, the pitch of the electron&#039;s helical path is approximately 0.0089 meters.

An electron enters a uniform magnetic field B

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